cJf,  (^j(Y^^' 


'^ 


ELEMENTS 


o;d 


PLANE    GEOMETRY. 


ELEMENTS 


OF 


PLANE   GEOMETRY. 


PAET   I. 


WITH  AN  APPENDIX  ON   MENSURATION. 


By  THOMAS  HUNTER,  A.M., 

PRESIDENT   OP  THE  NORMAL   COLLEGE   OF   THE   CITY   OF   NEW  YORK. 


NEW    YORK: 

HARPER  &   BROTHERS,  PUBLISHERS, 

FRANKLIN     SQUAKE. 

18  72. 


Entered  according  to  Act  of  Congress,  in  the  year  1871 ,  by 

Harper  ^  IBrothers, 

In  the  Office  of  the  Librarian  of  Congress,  at  Washington, 


5U, 


AJORl 


PREFACE. 


The  only  apology  necessary  for  adding  another  to 
the  many  text-books  with  which  the  market  is  over- 
stocked is  that  this  little  work  on  Geometry  is  very 
much  needed.  The  greater  portion  of  it  was  prepared 
while  engaged  in  teaching  the  subject,  and  while  all  the 
difficulties  of  presenting  it  in  a  practical  manner  were 
fresh  in  the  author's  mind.  Several  friends — them- 
selves superior  instructors — strongly  advised  its  pub- 
lication, in  the  hope  that  it  would  render  the  study  of 
the  elementary  principles  of  Geometry  more  simple  and 
easy  of  comprehension. 

In  teaching  Geometry,  it  appeared,  at  first,  singular 
that  students  could  master  the  difficulties  of  Arithmetic 
and  Algebra,  and  yet  fail  to  comprehend  the  relations 
of  magnitudes  which  appealed  to  the  sense  of  sight. 
A  little  closer  observation,  however,  revealed  the  fact 
that  many  students  accomplished  little  more  than  the 
committing  to  memory  variations  of  "A,"  "B,"  "C,'' 
and  "  1,"  "  2,"  "  3.'*  Of  correct  geometrical  reasoning 
they  had  hardly  a  conception.  Even  the  intelligent 
pupils  were  found  unable  to  apply  the  principles  to 
new  matter;  and  the  solution  of  problems  not  in  the 
book  was  almost  an  impossibility.  Geometry,  if  prop- 
erly taught  and  thoroughly  understood,  is  just  as  flex- 
ible as  Arithmetic  or  Algebra. 

Geometry  was  truly  a  "rope  of  sand''  to  whole 
classes.      The   necessary  examination   completed,   the 

318240 


VI  PREFACE. 

subject  was  abandoned  and  forgotten.  The  principal 
cause  for  this  was  a  wide  departure  by  many  recent 
writers  from  the  rigid  system  of  Euclid.  For  example, 
Euclid  commences  with  the  simple  problem,  ^'On  a 
given  straight  line  to  construct  an  equilateral  triangle." 
By  means  of  the  postulate  or  problem,  whose  solution 
is  selfevident,  that  *'A  circle  may  be  described  with 
any  point  as  centre  and  any  length  of  Jine  as  radius," 
how  simple,  beautiful,  and  satisfactory  to  the  mind  of 
the  learner  the  construction  becomes!  Besides,  the 
compasses  and  ruler  are  placed  in  the  hand  of  the 
student  from  the  very  beginning;  he  does  something 
for  himself;  sees  its  truth,  and  assimilates  it  with  his 
intellect.  Nearly  all  the  Geometries  in  use  in  the 
schools  commence  with  a  theorem.  The  pupils  are 
told  to  erect  a  perpendicular  from  a  given  point  in  a 
given  line.  By  what  authority?  By  a  postulate! 
Then  it  is  a  problem  whose  solution  is  self-evident! 
If  so,  why  after  using  it  to  establish,  link  by  link,  a 
chain  of  truths  extending  through  three  books,  does 
the  author  proceed  to  demonstrate  it?  It  reminds  us 
of  a  man  who,  building  a  superstructure  on  a  false 
foundation,  is  forced  to  pause  in  his  work  when  he 
has  completed  his  third  story,  and  reconstruct  a  true 
foundation  to  prevent  the  whole  edifice  from  toppling 
over.  It  is  a  problem  whose  solution  is  self-evident 
that  "A  line  may  be  bisected."  Why  not,  also,  that 
it  may  be  trisected?  Five  postulates  are  subsequently 
demonstrated.  First,  they  are  self-evident ;  second,  they 
are  not  self-evident,  and  require  solution!  Is  it  any 
wonder  that  the  youthful  mind  is  shocked  db  initio? 
Imagine  Euclid  asking  his  auditors:  ^'I  beg  that  you 
will  grant  that  a  straight  line  may  be  drawn  through 
a  given  point  parallel  to  a  given  line."  His  auditors 
would  have  granted  no  such  thing,  and  would  have 


PREFACE. 


told  him  he  was  begging  too  much.  It  is  really  more 
** self-evident"  that  "if  one  straight  line  cut  another 
straight  line,  the  opposite  or  vertical  angles  are  equal." 
Why  not  beg  this  too?  or,  indeed,  beg  the  whole  sub- 
ject? 

Three  other  works  on  Geometry  contain  no  postulates 
whatever !  If  Geometry  be  founded  on  definitions,  ax- 
ioms, and  postulates,  it  is  certainly  a  violation  of  rigid 
geometrical  reasoning  to  omit  any  necessary  step  in  the 
process.  If  the  first  fifteen  or  twenty'  propositions  are 
thoroughly  taught  and  perfectly  mastered,  the  subse- 
quent study  of  geometry  is  comparatively  simple.  The 
aim  of  the  teacher  should  be  to  train  the  scholar  rigidly 
— to  take  nothing  for  granted,  unless  really  self-evident 
or  previously  demonstrated.  The  lohy  and  the  where- 
fore of  every  step  must  be  stated.  Each  link  in  the 
chain  must  be  as  strong  as  any  other  link. 

A  great  evil  has  arisen  from  the  attempts  to  shorten 
the  demonstrations  of  the  propositions.  Important 
omissions  are  likely  to  occur,  and  haste  and  inaccu- 
racy frequently  follow.  It  is  much  better  to  make  the 
proof  complete  and  satisfactory,  so  that  the  student  will 
not  be  obliged  to  review  again  and  again.  In  every 
geometrical  demonstration  so  much  and  no  more  is 
necessary.  It  is  as  bad  to  omit  as  to  add ;  and  great 
care  should  be  exercised  in  giving  just  enough.  Cir- 
cumlocution is  tedious ;  but  lack  of  thoroughness  often 
vitiates  the  truth.  Besides  its  practical  utility,  the 
study  of  Geometry  imparts  a  love  for  truth  for  its 
own  sake ;  it  strengthens  the  reasoning  faculties  more 
than  the  study  of  any  of  the  other  mathematical  sci- 
ences; and,  unless  carried  to  too  great  an  extent,  cul- 
tivates clearness,  precision,  and  brevity  of  expression. 

The  present  volume  is  intended  only  for  beginners, 
for  those  who  are  preparing  for  college,  and  for  inter- 


Vlll  PREFACE. 


mediate  and  higli  schools  generally.  The  Geometry  of 
Planes  and  Solids  is  omitted.  Nearly  all  the  works 
hitherto  published  on  this  subject  contain  in  addition 
appendices  on  Plane  and  Spherical  Trigonometry  and 
Logarithms.  The  vast  majority  of  students  rarely  ad- 
vance beyond  the  geometry  of  lines,  angles,  and  plane 
figures.  The  work  in  its  present  form  will  be  cheap, 
and  will  exhaust  the  first  and  most  important  depart- 
ment of  geometrical  study.  Any  pupil  wishing  to 
make  further  progress  can  readily  do  so  by  taking  up 
any  other  work  on  the  subject.  Should  the  present 
volume,  however,  accomplish  its  mission,  the  author 
vjill  publish  a  second  volume  containing  the  higher 
departments. 

We  claim  for  this  little  work  on  Geometry — 1.  That 
it  commences  aright;  2.  That  it  contains  more  prob- 
lems, solved  and  unsolved,  than  any  other  volume  of 
its  size  extant;  3.  That  it  is  more  practical  than  the 
works  generally  in  use ;  4.  That  it  contains  an  appen- 
dix on  Mensuration  of  Surfaces,  which  furnishes  a  use- 
ful application  of  Arithmetic  to  the  Geometry  previ- 
ously studied. 


CONTENTS. 


BOOK  I. 

Page 

Explanation  of  Terms 11 

Hints  to  Teachers 18 

Lines,  Angles,  and  Triangles 20 

Test  Examples 44 

BOOK  II. 

Quadrilaterals,  Squares,  etc 48 

Test  Examples 63 

BOOK  III. 

The  Circle 70 

Test  Examples 82 

BOOK  IV. 

Proportion 89 

BOOK  V. 

Proportion  applied  to  Figures 95 

Test  Examples 114 


APPENDIX. 
Mensuration  of  Surfaces '. 125 


A* 


ELEMENTS  OF  PLANE  GEOMETRY. 


BOOK  I 

EXPLANATION    OF    TERMS. 

/  1.  Geometey  is  that  science  which  treats  of  the  rela- 
tion and  measurement  of  magnitudes. 
,,  2.  Magnitudes  have  three  dimensions — length,  breadth, 
and  thickness. 

3.  The  Science  of  Geometry  is  founded  on  Definitions, 
Axioms,  and  Postulates. 

4.  A  Definition  is  an  explanation  of  any  term  or  word 
in  a  science,  showing  the  sense  in  which  it  is  employed. 

,5.  An  Axiom  is  a  self-evident  truth. 

,6.  A  Postulate  is  a  self-evident  problem. 
^  7.  A  Theorem  is  a  proposition  requiring  a  demonstra- 
tion. 
^   8.  A  Problem  is  a  proposition  requiring  a  solution. 

9.  A  Demonstration  is  a  chain  of  logical  arguments 
establishing  the  truth  of  some  proposition. 
..   10.  A  Direct  or  Positive  Demonstration  is  one  that 
concludes  with  certain  proof  of  the  proposition. 

31.  An  Indirect  or  Negative  Demonstration  is  one 
which  proves  a  proposition  to  be  true  by  demonstrating 
that  some  absurdity  must  follow  if  the  proposition  ad- 
vanced were  false. 

12.  A  Lemma  is  a  preparatory  proposition  employed 
for  the  demonstration  .of  a  theorem  or  the  solution  of  a 
problem. 


12  ELEMENTS  OF  PLANE  GEOMETRY. 

13.  A  Corollary  is  an  obvious  consequence  deduced 
from  one  or  more  propositions. 

14.  A  Scholium  is  a  remark  on  one  or  more  preceding 
propositions,  showing  their  use,  their  connection,  their 
restriction,  or  their  extension. 

15.  An  Hypothesis  is  a  supposition  assumed  to  be  true 
in  the  statement  of  a  proposition. 

Signs,  • 

1.  The  sign  of  Equality  is  two  parallel  straight  lines 
of  equal  length ;  thus,  A=B  is  read  A  equals  B. 

2.  The  sign  of  Inequality  is  an  acute  angle:  the  great- 
er quantity  is  placed  at  the  opening  of  the  angle ;  thus, 
A>B  is  read  A  greater  than  B. 

3.  The  sign  of  Addition  is  an  erect  cross;  thus,  A+B 
is  read  A  plus  B,  and  means  that  B  is  added  to  A. 

4.  The  sign  of  Subtraction  is  a  horizontal  line ;  thus, 
A—B  is  read  A  minus  B,  and  means  that  B  is  to  be  taken 
from  A. 

5.  The  sign  of  Multiplication  is  an  oblique  cross  ;  thus, 
A  X  B  is  read  A  multiplied  by  B,  and  means  that  A  is  taken 
B  times.  It  is  also  expressed  by  a  point,  or  by  simply 
writing  the  letters  together;  thus,  A.B  or  AB. 

6.  All  the  quantities  w^ithin  parentheses,  braces,  or 
brackets,  or  under  a  vinculum,  are  considered  as  one 
quantity;    thus,    (A+B— C),    {A+B~C},   [A+B-C], 

A+B-0. 

7.  The  sign  of  Division  is  a  horizontal  line,  with  a  dot 
above  and  another  below;  thus,  A-f-B  is  read  A  divided 
by  B ;  or  the  division  is  expressed  by  making  A  the  nu- 

A 

merator  of  a  fraction,  and  B  the  denominator;  thus,  ~. 

8.  The  power  of  a  quantity  is  expressed  by  means  of  a 
figure  or  letter  placed  to  the  right  and  a  little  above  the 
quantity ;  thus,  A^,  A^,  A*,  A"*,  is  read  A  squared,  A  cubed, 
A  raised  to  the  fourth  power,  A  raised  to  the  ^nth  power. 
2,  3,  4,  and  in  are  called  exponents,  or  indices. 


liOOK    I.  13 

9.  The  root  of  a  quantity  is  expressed  by  means  of  a 
symbol  called  the  radical  sign,  with  a  figure  or  letter  to 
indicate  the  particular  root ;  thus,  -y/A^  -y^A,  -^A,  ^A  is 
read  the  square  root,  the  cube  root,  the  fourth  root,  the 
mill  root  of  A ;  or  these  roots  may  be  expressed  by  frac- 
tional exponents ;  thus  A^,  AJ,  A^,  and  A^. 

10.  The  sign  oi  therefore^  or  hence^  is  three  dots  placed 
in  a  triangular  form ;  thus  /. 

11.  A  Ratio  is  a  quotient;  the  ratio  of  3  to  4  is  -. 

12.  A  Proportion  is  an  equality  of  Ratios;  and  is  writ- 

A    C 
ten  thus:  tt=t^,  A-t-B=:C-~D,  or  A:B::C:D,  and  is 

read  A  is  to  B  as  C  to  D. 

Definitions, 
.     1.  A  Point  is  that  which  has  position  only. 
,    2.  A  Line  is  length  without  breadth. 

3.  A  Straight  Line  is  one  that  does  not  change  its  di- 
/^    rection  at  any  point,  or  it  is  the  shortest  distance  be- 
tween two  points.     A  straight  line  can  not  include  a 
space  or  a  segment. 

4.  A  Curved  Line  is  one  that  changes  its  direction  at 
^  every  point. 

,  J        5.  A  Broken  Line  is  made  up  of  two  or  more  straight 
lines  not  lying  in  the  same  direction. 
f      6.  A  Surface,  or  Superficies,  is  that  which  has  length 
and  breadth. 

y     7.  A  Solid  is  that  which  has  length,  breadth,  and  thick- 
ness. 

/  8.  The  boundaries  of  solids  are  surfaces;  of  surfaces, 
''  lines ;  and  the  extremities  of  lines,  points.  Imagine  a 
point  moved  forward :  it  would  generate  a  line ;  the  line 
moved  forward  would  generate  a  surface ;  and  the  sur- 
face moved  forward  would  generate  a  solid. 

9.  An  Angle  is  the  difference  of  direction  of  two  straight 
lines  meeting  in  a  point. 


14  ELEMENTS  OF  PLANE  GEOMETBY. 

10.  When  a  straight  line  meets  another 
straight  line,  and  makes  the  adjacent  an- 
gles equal,  each  angle  is  called  a  right 
angle ;  and  the  line  which  meets  the  oth- 
er is  said  to  be  a  perpendicular  to  it. 
I 

11.  An  Acute  Angle  is  an  angle 
less  than  a  right  angle. 


12.  An  Obtuse  Angle  is  an  an- 
gle greater  than  a  right  angle. 


13.  A  Plane  Figure  is  a  portion  of  a  surface  bounded 
by  straight  or  curved  lines. 

14.  The  area  of  a  figure  is  the  quantity  of  space  con- 
tained in  it. 

1 5.  A  Circle  is  a  plane  figure  bound- 
ed by  a  curved  line  such  that  every 

j,  point  upon  it  is  equally  distant  from  a 
point  within  it  called  the  centre. 

16.  The  curved  line  which  bounds 
it  is  called  the  Circumference. 

1 7.  The  Diameter  of  a  circle  is  a  straight  line  passing 
through  the  centre,  and  terminating  both  ways  in  the 
circumference. 

18.  A  Radius  is  a  straight  line  drawn  from  the  centre 
to  any  part  of  the  circumference  of  a  circle. 

1 9.  AH  Radii  of  the  same  circle  are  equal ;  all  diameters 
of  the  same  circle  are  also  equal ;  and  each  diameter  is 
double  the  radius. 

20.  A  Polygon  is  a  plane  figure  bounded  by  straight 
lines;  these  lines  are  called  sides,  and  the  broken  line 
forming  the  boundary  is  called  the  perimeter. 

21.  A  Triangle  is  a  plane  figure  bounded  by  three 
straight  lines.     Th^re  are  six  sorts  of  triangles;  tJiree 


BOOK   I. 


15 


with  respect  to  their  sides,  and  three  with  respect  to 
their  angles.  The  three  with  respect  to  their  sides  are 
equilateral,*  isosceles,f  and  scalene  ;l  and  the  three  with 
respect  to  their  angles  are  right-angled,  acute-angled,  and 
obtuse-anorled. 


22.  An  Equilateral  Triangle  is  a  tri- 
angle that  has  its  three  sides  equal. 


23.  An  Isosceles  Triangle  is  a  triangle  that 
has  only  two  of  its  sides  equal. 


24.  A  Scalene  Triangle  is  a  triangle  that 
has  no  two  of  its  sides  equal. 


25.  A  Right-angled  Triangle  is  one  that 
contains  a  right  angle. 


*  From  aequus,  equal,  and  latus,  a  side ;  equal-sided. 
t  From  two  Greek  words  meaning  equal  legs ;  hence  a  triangle  having 
two  equal  sides. 

t  Scalene  means  squinting. 


ELEMENTS  OF  TLANE  GEOMETKY. 


26.  An  Acute-angled  Triangle  is  one  that 
has  all  its  angles  acute. 


27.  An  Obtuse-angled  Triangle  is  one 
that  has  an  obtuse  ansrle. 


28.  Parallel  lines  are  such  as  lie  in  the 
same  plane,  and  can  not  meet,  how  far  so- 
ever they  may  be  produced  either  way. 
29.  A  Quadrilateral,  a  Quadrangle, 
or  a  Trapezium,  is  a  plane  figure  of 
four  sides. 

/       30.  A  Parallelogram  is  a  quadrilateral 
whose  opposite  sides  are  parallel. 

31.  A  Trapezoid  is  a  quadrilateral  which 
has  only  two  of  its  sides  parallel. 

32.  A  Rhomboid  is  a  parallelogram  that  has  no  right 
angle. 


33.  A  Rectangle  is  a  right-angled  par- 
allelogram. 


34.  A  Square  is  an  equilateral  rectangle. 


BOOK   I.  17 

35.  A  Rhombus  is  an  equilateral  rhom- 
boid. 

36.  A  Diagonal  is  a  straight  line  con- 
necting two  angles  not  consecutive. 

37.  The  Hypothenuse  of  a  right-angled  triangle  is  the 
side  opposite  the  right  angle. 

38.  The  Base  of  a  plane  figure  is  that  side  upon  which 
it  is  supposed  to  stand.  Any  side  may  be  considered  the 
base. 

Axioms, 

1.  Things  which  are  equal  to  the  same  thing  arc  equal 
to  each  other. 

2.  If  equals  be  added  to  equals  the  sums  are  equals. 

3.  If  equals  be  taken  from  equals  the  remainders  are 
equals. 

4.  If  equals  be  added  to  unequals  the  sums  will  be  un- 
equals. 

5.  If  equals  be  subtracted  from  unequals  the  remain- 
ders will  be  unequals. 

C.  Things  which  are  double  of  the  same  or  equal  things 
are  equal. 

7.  Things  which  are  halves  of  the  same  or  equal  things 
are  equal. 

8.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

9.  The  whole  is  greater  than  any  one  of  its  parts. 

10.  Things  which  coincide,  or  fill  the  same  space,  are 
identical  and  equal  in  all  their  parts. 

11.  All  right  angles  are  equal. 

12.  Two  straight  lines  which  intersect  each  other  can 
not  both  be  parallel  to  the  same  straight  line. 

/  Postulates. 

1.  Let  it  be  granted  that  a  straight  line  may  be  drawn 
from  one  point  to  another ; 

2.  That  a  straight  line  may  be  produced  to  any  length ; 


18  ELEMENTS    OF   PLANE    GEOMETKY. 

3.  And  that  a  circle  may  be  described  from  any  point 
as  centre,  and  with  any  length  of  line  as  radius. 

HINTS  TO  TEACHERS. 

Before  commencing  the  study  of  the  Propositions  of 
Geometry  it  is  absolutely  necessary  that  the  students 
should  be  provided  with  compasses  and  rulers.  They 
should  draw  lines  of  all  sorts  and  sizes;  they  should 
place  them  in  such  a  manner  as  to  form  all  kinds  of  an- 
gles, triangles,  quadrilaterals ;  and  they  should  write  the 
names  of  the  different  figures.  The  students  must  com- 
mence with  measuring ;  otherwise  they  are  not  studying 
Geometry  ;  they  are  simply  committing  to  memory  varia- 
tions of  A,  B,  C  and  1,  2,  3.  One  of  the  chief  difficulties 
encountered  hitherto  in  the  study  of  this  beautiful  sci- 
ence arose  from  the  fact  that  the  text-books  in  common 
use  begin  with  a  theorem^  and  that  theorem  none  of  the 
easiest.  In  the  present  work  the  first  proposition  is  a 
prohlem.  But  although  this  is  the  case  the  scholars 
should  be  compelled  to  draAV  accurately  and  neatly  the 
following : 

1.  With  ruler  and  pen  draw  one  inch*  on  the  horizontal. f 

2.  Draw  four,  six,  eight,  and  ten  inches  on  the  hori- 
zontal. 

3.  Draw  four,  six,  eight,  and  ten  inches  on  the  vertical.f 

4.  Draw  a  line  equal  to  the  sum  of  two  other  lines. 

5.  Draw  a  line  equal  to  the  sum  of  four  other  lines. 

6.  Draw  a  line  equal  to  two  other  lines  of  four  inches 
each. 

7.  Draw  a  horizontal  line  equal  to  two  inches,  and  an 

*  If  the  student  does  not  thoroughly  comprehend  distance,  the  inch, 
the  foot,  the  yard,  the  pole,  the  mile,  the  league  should  be  carefully 
taught. 

t  The  horizon  is  the  line  that  bounds  our  view.  Hence  a  horizontal 
line  is  a  line  parallel  with  the  horizon.  A  vertical  line  is  perpendicular 
to  the  horizon. 


BOOK    I.  19 

oblique  line  touching  its  extremity.     Cut  off  from  the 
oblique  line  two  inches. 

8.  Draw  a  vertical  line  six  inches  long,  and  from  the  low- 
er extremity  draw  a  horizontal  line  of  the  same  length. 

9.  Describe  a  circle  with  any  point  as  centre,  and  any 
length  of  line  as  radius. 

10.  Describe  a  circle  with  a  radius  equal  to  two  inches. 
^1.  Draw  a  radius,  a  diameter,  a  chord.* 

12.  Draw  a  tangent,f  a  secant. J 

13.  Draw  a  triangle,  a  quadrilateral,  a  pentagon.§ 

14.  Draw  all  kinds  of  angles. 

15.  Draw  two  diameters  at  right  angles,||  and  join  their 
extremities. 

,16.  Draw  four  tangents  at  right  angles,  and  join  their 
extremities. 

'  1 7.  Lay  off  the  radius  on  the  circumference  six  tim^s, 
and  join  the  points;  join  the  alternate  points.  (What 
figures  have  you  formed  ?) 

18.  Draw  a  parallelogram,  a  rectangle,  a  rhomboid.. 

19.  Make  a  square,  and  divide  it  into  four  equal  parts. 

20.  Divide  a  circle  into  four  equal  parts. 

21.  Draw  a  rectangle  twice  as  long  as  it  is  broad. 

.22.  Divide  a  hexagon^  into  six  parts  by  drawing  lines 
from  a  point  within  it. 

23.  Make  a  right-angled  triangle,  each  of  whose  sides 
shall  be  double  the  sides  of  a  given  right-angled  triangle. 

24.  Make  a  square,  each  side  of  which  shall  be  three 
times  one  inch. 

*  A  chord  is  a  line  which  cuts  off  a  portion  of  a  circle ;  it  terminates 
in  the  circumference  both  ways. 

t  A  tangent  is  a  line  which  touches  the  circumference  of  a  circle. 

X  A  secant  is  a  line  which  cuts  the  circumference. 

§  A  pentagon  is  a  figure  of  five  sides. 

II  Of  course,  the  leanier  performs  these  problems  to  the  best  of  his 
ability.  The  object  is  simply  to  make  him  expert  in  the  use  of  instru- 
ments, and  to  fasten  on  his  mind  more  thoroughly  the  definitions  already 
committed  to  memory. 

H  A  hexagon  is  a  figure  of  six  sides. 


1^ 


20 


ELEMENTS  OF  TLAXE  GEOMETRY. 


Pkoposition  I. — Problem. 

On  a  given  straight  li^ie  to  construct  an  equilateral  tri- 
angle. 

Let  AB  be  a  given  straight 
line;  it  is  required  to  con- 
struct upon  it  an  equilateral 
triangle. 

From  the  point  A  as  centre, 
and  with  AB  as  radius  (Postu- 
late 3),  describe  the  circle 
BCD ;  and  from  the  point  B  as 
centre,  and  with  AB  as  radius,  describe  the  circle  ACE ; 
and  from  the  point  C,  where  the  circles  cut  each  other, 
draw  the  straight  lines  (Post.  1)  AC  and  BC  to  the  points 
A  and  B ;  ABC  is  an  equilateral  triangle. 

Because  the  point  A  is  the  centre  of  the  circle  BCD, 
AC  is  equal  to  AB  (Definition  19),  and  because  the  point 
B  is  the  centre  of  the  circle  ACE,  BC  is  equal  to  AB. 
Hence,  since  AC  and  CB  are  each  equal  to  AB,  they  must 
be  equal  to  each  other  (Axiom  1) ;  therefore  AB,  AC.  and 
CB  are  equal,  and  the  triangle  ABC  is  equilateral. 

PRorosiTioN  II. — Problem. 

From  a  given  point  to  draw  a  straight  line  equal  to  a 
given  straight  line. 

Let  A  be  the  given  point  and  BC 
the  given  line;  it  is  required  to 
draw  from  the  point  A  a  straight 
line  equal  to  BC. 

From  the  point  A  to  B  draw  the 
line  AB  (Post.  1),  and  upon  it  con- 
struct an  equilateral  triangle,  ABF 
(Proposition  I.) ;  from  the  point  B 
as  centre,  and  with  the  radius  BC, 
describe  the  circle  GCL  (Post.  3) ;  produce  the  line  FB 
until  it  meets  the  circumference  at  G;    then,  from  the 


BOOK    I.  21 

point  F  as  centre,  and  with  tlic  line  FG  as  radius,  de- 
scribe the  circle  GEH;  produce  FA  until  it  meets  the 
circumference  at  E.     The  line  AE  will  be  equal  to  BC. 

Because-  the  point  B  is  the  centre  of  the  circle  GCL, 
BC  is  equal  to  BG ;  and  because  F  is  the  centre  of  the 
circle  GEH,  FG  is  equal  to  FE  (Def.  19) ;  but  BF  and 
AF  are  equal;  therefore  (Ax.  3),  if  these  equals  be  sub- 
tracted, the  remainders,  AE  and  BG,  are  equal ;  but  it  was 
before  proved  that  CB  is  equal  to  BG,  and  since  AE  and 
BC  are  each  equal  to  BG,  they  must  be  equal  to  each 
other. 

PROPOSITION  III. — Problem. 

From  the  greater  of  two  given  straight  lines  to  cut  off  a 
part  equal  to  the  less. 

Let  AB  and  D  be  two  given  straight  lines,  of  which 
AB  is  the  greater;  it  is  required  to  cut  oiF  from  AB  a 
part  which  shall  be  equal  to  D. 

From  the  point  A  draw 
the  line  AE  equal  to  D 
(Prop.  II.) ;  and  from  the 
centre  A,  and  with  the  ra- 
dius AE, -describe  the  cir- 
cle ECF  (Post.  3) ;  then  will 
AC  be  equal  to  D. 

The  line  AE  is  equal  to 
the  line  D  by  construction,  and  the  line  AC  is  equal  to 
the  line  AE  (Def.  19).  Now,  since  AC  and  D  are  each 
equal  to  AE,  they  must  be  equal  to  each  other  (Ax.  1). 

Pkopositiox  IV. — Theorem. 

If  tioo  triangles  have  two  sides  and  the  included  angle 
of  the  one  equal  to  tioo  sides  and  the  included  angle  of  the 
other ^  each  to  each^  the  triangles  are  every  way  equal 

Let  ACB  and  DFE  be  two  triangles  which  have  the 
two  sides  AC  and  CB  and  the  included  angle  ACB  of 
the  one  equal  to  DF  and  FE  and  the  included  angle 


22  ELEME^^TS    OF    PLAXE    GE05IETKY. 

DFE  of  the  other; 
then  will  the  two 
triangles  ACB  and 
DFE  be  every  way 
equal. 

For  let  the  trian- 
■^  gle  ACB  be  applied 
to  the  triangle  DFE,  so  that  the  point  C  may  fall  on 
the  point  F,  and  the  line  AC  upon  DF ;  and  since  AC 
is  equal  to  DF  the  point  A  must  coincide  with  the  point 
D ;  since  AC  coincides  with  DF,  and  the  angle  ACB  is 
equal  to  the  angle  DFE,  the  line  CB  must  coincide  with 
the  line  FE;  and  since  CB  is  equal  to  FE  the  points 
B  and  E  must  coincide ;  since  the  points  A  and  D  coin- 
cide, and  the  points  B  and  E  coincide,  the  line  AB  must 
coincide  with  the  line  DE  (Ax.  10);  therefore  the  two 
triangles  coincide  throughout  their  whole  extent,  and 
are  every  way  equal  (Ax.  10). 

Peopositiox  Y. — Theorem. 

If  two  triangles  ham  two  angles  and  the  included  side 
of  the  one  equal  to  two  angles  and  the  included  side  of  the 
other ^  each  to  each^  the  triangles  are  every  way  equal. 

Let  the  two  tri- 
angles ABC  and 
DEF  have  the  an- 
gles A  and  B  equal 
to  the  angles  D  and 
E,  each  to  each,  ancj  the  included  side  AB  equal  to  the 
included  side  DE,  then  will  the  two  triangles  be  every 
way  equal. 

For  apply  the  triangle  ABC  to  the  triangle  DEF,  in 
such  a  manner  that  the  line  AB  will  coincide  with  the 
line  DE,  and  since  the  angle  A  is  equal  to  the  angle  D, 
the  side  AC  will  take  the  direction  DF,  and  the  point  C 
will  fall  somewhere  on  the  line  DF ;  and  since  the  angle 


BOOK    I.  23 

B  is  equal  to  the  angle  E,  the  side  BC  must  take  the  di- 
rection of  EF,  and  the  point  C  must  fall  somewhere  on 
the  line  EF.  Now,  since  the  point  C  falls  at  the  same 
time  upon  the  lines  DF  and  EF,  and  since  it  can  not  fall 
in  two  places  at  the  same  time,  it  must  fall  at  their  point 
of  intersection,  F.  Therefore  the  points  C  and  F  coincide, 
and  the  two  triangles  coincide  thi*oughout  their  whole  ex- 
tent, and  are  therefore  every  way  equal  (Ax.  10).  , 

Peopositiox  yi. — Theorem. 

The  angles  at  the  base  of  an  isosceles  triangle  are  eqnal^ 
and^  if  the  equal  sides  he  produced^  the  angles  below  the 
base  are  also  eqitaL 

Let  ABC  be  an  isosceles  triangle ; 
then  will  the  angles  CAB  and  CBA 
be  equal,  and  also  the  angles  BAD 
and  ABE  ^he  equal  sides  being  pro- 
duced). 

From  AF  cut  off  any  part  AD,  and 
from  BG  cut  off  a  23art,  BE,  equal  to 
AD;  join  the  points  A  and  E,  and  D 
and  B. 

The  lines  AC  and  BC  are  equal 
(hypothesis),  and  AD  and  BE  (con- 
struction) are  also  equal.  To  AC  add  AD,  and  to  BC 
add  BE,  and  CD  will  be  equal  to  CE  (Ax.  2).  In  the 
tw'o  triangles  DCB  and  ECxA.,  the  sides  DC  and  CB  are 
equal  to  the  sides  EC  and  CA,  each  to  each,  and  the 
angle  C  is  common  to  both.  Hence  the  tw^o  triangles 
are  every  way  equal  (Prop.  ,IY.)  ;  the  angle  CDB  is  equal 
to  CEA,  and  CBD  is  equal  to  CAE,  and  the  side  DB  is 
equal  to  the  side  EA.  Again,  in  the  tw^o  triangles  ADB 
and  BEA,  the  sides  AD  and  BE  are  equal ;  and  it  was 
before  proved  that  DB  and  EA  are  equal,  and  that  the 
angle  ADB  is  equal  to  the  angle  BEA.  Hence  the  two 
triangles  have  two  sides  and  the  included  angle  of  the 


24 


ELEMENTS    OF    PLANE    GEOMETRY. 


one  equal  to  two  sides  and  the  included  angle  of  the 
other,  each  to  each ;  they  are,  therefore,  equal  in  all  their 
parts  (Prop.  lY.) ;  the  angle  DAB  is  equal  to  the  EBA, 
these  are  the  angles  below  the  base,  and  the  angle  DBA 
is  equal  to  the  angle  EAB ;  but  it  was  before  proved  that 
DBC  is  equal  to  EAC ;  if  from  these  equals  the  equals 
DBA  and  EAB  be  taken,  there  will  remain  (Ax.  3)  the 
angle  ABC,  equal  to  the  angle  BAG ;  these  are  the  angles 
at  the  base. 


Proposition  VII. — Theorem. 


A 


If  two  triangles  have  three  sides  of  the  one  equal  to 
three  sides  of  the  other ^  each  to  each^  the  triangles  icill  he 
equal  iii  all  their  parts. 

Let  the  two  trian- 
gles ABC  and  DEF 
have  the  sides  AB, 
AC,  and  BC  equal 
to  the  sides  DE,  DF, 
and  EF,  each  to 
each  ;  then  will  these 
triangles  be  equal  in 
all  their  parts. 
Conceive  the  triangle  ABC  applied  to  the  triangle 
DEF,  so  that  the  equal  sides  AB  and  DE  will  coincide, 
and  the  sides  AC  and  BC  will  take  the  direction  of  DG 
and  EG.  Join  the  points  F  ^nd  G.  The  sides  DF  and 
DG  are  equal  (hyp.),  and  the  triangle  FDG  is  isosceles; 
hence  the  angle  DFG  is  equal  to  the  angle  DGF  (Prop. 
VI.).  In  like  manner  it  can  be  demonstrated  that  the 
angle  EFG  is  equal  to  EGF.  Therefore  (adding  these 
equals)  the  angle  DFE  is  equal  to  the  DGE  (Ax.  2).  In 
the  two  triangles  DFE  and  DGE,  the  sides  DF  and  FE 
are  equal  to  DG  and  GE,  each  to  each  (hyp.),  and  the 
included  angles  have  just  been  proved  equal;  hence  the 
triangles  are  equal  in  all  their  parts  (Prop.  IV.) ;  but  the 


BOOK   I. 


25 


triangle  DGE  is  equal  to  the  triangle  ABC.     Therefore 
ABC  and  DFE  are  equal  in  all  their  parts. 

Proposition  VIII. — Problem. 
To  bisect  a  given  angle. 

Let  BAC  be  the  given 
angle ;  it  is  required  to 
bisect  it — that  is,  to  di- 
vide it  into  two  equal 
parts. 

Take  any  point  D  on 
the  line  AB,  and  from 
AC  cut  off  a  part  AE  equal  to  AD  (Prop.  III.) ;  join  the 
points  E  and  D ;  and  on  the  line  ED  construct  an  equi- 
lateral triangle,  EDF ;  join  the  points  A  and  F ;  the  line 
AF  bisects  the  angle  BAC. 

In  the  two  triangles  ADF  and  AEF  the  sides  AD  and 
AE  are  equal  (const.),  FD  and  EF  are  also  equal  (Prop. 
I.),  and  the  side  AF  is  common  to  both ;  hence  (Prop. 
VII.)  the  two  triangles  are  every  way  equal,  and  the 
angle  DAF  is  equal  to  the  angle  EAF. 

Scholium,  By  the  same  construction  DAF  and  EAF 
may  be  divided  into  two  equal  parts ;  and  thus,  by  suc- 
cessive divisions,  a  given  angle  may  be  divided  into  four 
equal  parts,  into  eight,  into  sixteen,  and  so  on. 

Proposition^  IX. — Problem. 
To  bisect  a  given  straight  line. 
Let  AB  be  a  given  straight  line ;  it 
is  required  to  bisect  it. 

On  AB  construct  an  equilateral  tri- 
angle (Prop.  I.) ;  bisect  the  angle  ADB 
(Prop.  VIII.) ;  the  line  DC  will  bisect 
the  line  AB  at  the  point  C. 

In  the  two  triangles  ACD  and  BCD, 
the  sides  AD  and  BD  are  equal,  being  sides  of  an  equi- 
latei-al  triangle;  DC  is  common  to  bT>th,  and  the  angle 

B 


2G 


ELEMENTS    OF    PLxVXE    GEOMETRY. 


ADC  is  equal  to  the  angle  BDC  (const.).  Hence  the 
two  triangles  are  equal  in  all  their  parts  (Prop.  lY.),  and 
the  side  AC  is  equal  to  the  side  BC. 

r 

Pkoposition  X. — Problem. 

From  a  poiyit  on  a  straight  line  to  erect  a  perpendicular 
to  the  Ihie. 

Let  AB  be  a  given  line  and  C 
any  point  on  it ;  it  is  required 
to  erect  a  perpendicular  from 
this  point. 

Take  any  point,  D,  in  AC  and 
make  (Prop.  III.)  CE  equal  to 
DC ;  and  upon  DE  construct  an 
equilateral  triangle,  DFE,  and  join  F  and  C ;  the  line  FC 
is  perpendicular  to  AB. 

In  the  two  triangles  DFC  and  EFC,  DF  is  equal  to 
FE,  DC  to  CE  (const.),  and  FC  is  common;  hence  they 
have  three  sides  of  the  one  equal  to  three  sides  of  the 
other,  each  to  each.  Therefore  the  angle  DCF  (Prop. 
YII.)  is  equal  to  the  angle  ECF ;  and,  since  these  angles 
are  equal,  each  is  a  right  angle  (Def.  9).  Hence  CF  is 
perpendicular  to  AB. 

Proposition  XI. — Problem. 

From  a  given  point  without  a  straight  line  to  let  fall  a 
perpendiciilar  to  the  line. 

Let  AB  be  a  given  line  and  C  a 
point  without  it;  it  is  required  to 
draw  a  perpendicular  line  from  C 
to  AB. 

Take  any  point  F  on  the  opposite 
side  of  the  given  line,  and  from  the 
centre  C,  with  the  distance  CF  as 
radius,  describe  (Post.  3)  the  circle 
KFG,  meeting  AB  in  the  points  E  and  G.  Join  E  and 
C,  and  G  and  C ;  bisect  (Prop.  IX.)  the  line  EG  at  the 


BOOK    I. 


27 


point  H ;  join  II  and  C.     The  line  HC  will  be  perpendic- 
ular to  AB. 

.  In  the  two  triangles  ECH  and  GCH,  EC  and  GC  are 
equal  (Def.  15),  EH  and  HG  are  also  equal  (const.),  and 
HC  is  comnron  to  both.  Hence  the  two  triangles  are 
every  way  equal  (Prop.  YIL),  and  the  angle  EHC  is 
equal  to  the  angle  GHC;  and,  since  these  angles  are 
equal,  each  is  a  right  angle  (Def.  10).  Therefore  the  line 
HC  is  perpendicular  to  AB. 

Proposition  XII. — Theorem. 

Any  tioo  sides  of  a  triangle  are  together  greater  than 
the  third f^ 

Let  ABC  be  a  triangle ;  then  will 
the  sum  of  the  sides  AC  and  CB  be 
greater  than  AB. 

For  the  straight  line  AB  (Def.  2) 
is  the  shortest  distance  between  the 
two  points  A  and  B.  Hence  the 
broken  line  ACB  is  greater  than  the  straight  line  AB. 

Proposition  XIH. — Problem. 

Given  three  straight  lines^  the  sum  of  any  two  heing 
greater  than  the  third^\  to  construct  a  triangle  whose  three 
sides  shall  he  equal  to  the  given  lines^  each  to  each. 

Let  a^  h^  and  c  be 
the  given  straight 
lines;  it  is  required 
to  construct  a  trian- 
gle whose  three  sides 
shall  be  equal  to  these 
three  lines. 

Draw  an  indefinite 

*  This  proposition  is  an  obvious  deduction  from  the  definitions  of  a 
straight  line  and  of  a  triangle. 

t  A  straight  line  is  the  shortest  distance  between  two  points ;  hence, 
to  form  a  triangle,  the  sum  of  two  sides  must  be  greater  than  the  third. 


28  ELEMENTS  OF  PLANE  GEOMETRY. 

line,  DE,  and  on  it  lay  off  DF  equal  to  a,  FG  equal  to  b^ 
and  GK  equal  to  c;  then,  from  the  point  F  as  centre,  and 
with  the  line  DF  as  radius,  describe  the  circle  DHM  (Post. 
3),  and  from  the  point  G  as  centre,  and  with  the  line  GK 
as  radius,  describe  the  circle  HKM  (Post.  3).  Join  the 
points  F  and  G  with  the  point  of  intersection,  H ;  then 
will  FGH  be  the  required  triangle. 

The  line  a  is  equal  to  DF  (const.),  but  FH  is  equal  to 
DF  (Def  19);  therefore  FH  is  equal  to  a  (Ax.  1);  c  is 
equal  to  GK  (const.),  but  GH  is  equal  to  GK;  therefore 
GH  is  equal  to  c  (Ax.  1),  and  the  line  FG  is  equal  to  b 
(const.).  Hence  the  three  sides  of  the  triangle  are  equal 
to  the  three  given  lines. 

Proposition  XIY. — ^Problem. 

At  a  given  point  in  a  given  line  to  make  an  arigU 
equal  to  a  given  angle. 

£  ^      Let  B  be    the    given 

/]  A    point  in  the  straight  line, 

/  I  /  \    AB ;    it   is    required   to 

/      j  /     \    construct  an  angle  at  B 

X         j  /        \     equal  to  the  given  angle 

/  I  /  j     DEF. 

^ '^         y- ""^       Join  the  points  D  and 

^^  F,  and  construct  a  trian- 

gle GBC  (Prop.  Xin.),  which  shall  be  equal  to  the  trian- 
gle DEF,  having  the  side  GB  equal  to  DE,  BC  equal  to 
EF,  and  GC  equal  to  DF.  Hence  (Prop.  VII.)  the  angle 
GBC  is  equal  to  DEF. 

Proposition  XV. — ^Theorem. 

If  one  straight  line  meet  another  straight  line  {not  at 
the  extremity)  the  sum  of  the  angles  thus  formed  is  equal 
to  two  right  angles. 

Let  the  line  AB  meet  the  line  CD  at  the  point  C ;  then 
will  the  sum  of  the  angles  ACD  and  DCB  be  equal  to  two 
right  angles. 


BOOK  I.  29 

At  the  point  C  (Prop.  X.)  erect  the 
perpendicular  CE.  The  sjim  of  the 
angles  ACE  and  ECB  is  equal  to 
two  right  angles  (Def.  10).  But  the 
sum  of  the  angles  BCD,  DCE,  and  ^ — 
ACE  is  equal  to  the  sum  of  the  an- 
gles ACE  and  ECB.  Therefore  the  sum  of  the  angles 
BCD,  DCE,  and  ACE  is  equal  to  two  right  angles  (Ax. 
1).  But  the  angles  DCE  and  ACE  are  together  equal  to 
ACD  (Ax.  8) ;  therefore  the  sum  of  the  angles  BCD  and 
ACD  is  equal  to  two  right  angles. 

Peoposition  XVI. — Theorem. 

If  two  straight  lines  meet  a  third  straight  line,  making 
the  Stan  of  the  adjacent  angles  equal  to  two  right  angles, 
these  two  straight  lines  will  form  one  and  the  same  straight 
li?ie. 

Let  two  straight  lines,  AB  and  BC,  j> 

meet  a  third  straight  line,  BD,  mak-  y^ 

ing  the  sum  of  the  adjacent  angles,  y^ 

ABD  and  CBD,  equal  to  two  right  /  

angles,  then  will  these  two  lines  form  A  B 

one  and  the  same  straight  line. 

For,  suppose  that  AB  and  BC  do  not  form  one  and  the 
same  straight  line,*  then  some  other  line,  as  BE,  must  be 
the  continuation  of  AB;  then,  if  ABE  be  a  straight  line, 
the  sum  of  the  angles  ABD  and  EBD  is  equal  to  two 
right  angles  (Prop.  XV.).  But  (by  hypothesis)  the  sum 
of  the  angles  ABD  and  CBD  is  also  equal  to  two  right 
angles.  Hence  the  sum  of  the  angles  ABD  and  EBD  is 
equal  to  the  sum  of  the  angles  ABD  and  CBD ;  but  the 
angle  ABD  is  common  to  both ;  take  it  aw^ay,  and  there 
will  remain  the  angle  EBD  equal  to  CBD,  which  is  ab- 
surd (Ax.  9).      Therefore  BE  is  not  a  continuation  of 

*  This  method  of  demonstration  is  called  indirect,  or  the  reductio  ad 
ahsurdum. 


30  ELEMENTS  OP  PLANE  GEOMETRY. 

AB,  and  in  a  similar  manner  it  can  be  demonstrated  that 
no  other  line  than  BC  can  be  the  continuation  of  AB. 

Proposition  XYII. — Theorem. 

If  one  straight  line  intersect  another  straight  line^  the 
opposite  or  vertical  angles  are  equal, 

A^  ^        Let  the  straight  line  AB  inter- 

sect the  straight  line  CD  in  the 
point  G ;  then  will  the  angle  AGD 
be  equal  to  the  angle  CGB,  and 
AGO  equal  to  BGD. 
For  the  sum  of  the  angles  AGD  and  AGO  is  equal 
to  two  right  angles  (Prop.  XV.),  and  the  sum  of  the 
angles  BGC  and  AGO  is  also  equal  to  two  right  angles 
(Prop.  XIV.).  Therefore  the  sum  of  the  angles  AGD 
and  AGO  is  equal  to  the  sum  of  the  angles  BGC  and 
AGC.  Take  away  the  common  angle  AGC  (Ax.  3),  and 
the  remaining  angle  AGD  will  be  equal  to  the  remaining 
angle  BGC,  and  in  the  same  manner  it  can  be  demon- 
strated that  the  angle  AGC  is  equal  to  the  angle  BGD. 

Corollary.  It  is  manifest  that  the  sum  of  all  the  an- 
gles made  by  straight  lines  meeting  at  a  common  point 
is  equal  to  four  right  angles. 

Proposition  XVIII. — Theorem. 

If  one  side  of  a  triangle  he  produced^  the  exterior  angle 
is  greater  than  either  of  the  interior  and  <^§p09^  angles. 

Let  the  triangle  ABC  have 
one  of  its  sides,  AB,  produced 
to  D;  then  will  the  exterior 
angle  CBD  be  greater  than 
either  of  the  two  interior  and 
opposite  angles,  CAB  and  BCA. 
Bisect  BC  at  the  point  F 
(Prop.  IX.) ;  join  the  points  F 

and  A ;   produce  AF  until  FE  is   equal  to  it ;  join  B 

and  E. 


BOOK    I.  81 

In  the  two  triangles  AFC  and  EFB,  the  sides  CF  and 
FB  are  equal  (const.),  the  sides  AF  and  FE  are  also  equal 
(const.),  and  the  angles  AFC  and  EFB  are  equal  (Prop. 
XVII.).  Therefore  the  two  triangles  have  two  sides 
and  the  included  angle  of  the  one  equal  to  two  sides 
and  the  included  angle  of  the  other,  each  to  each,  and 
are  every  way  equal  (Prop.  IV.) ;  the  angle  EBF  is  equal 
to  the  angle  FCA.  But  DBC  is  greater  than  EBF  (Ax. 
9),  and  therefore  greater  than  FCA.  In  the  same  man- 
ner, by  producing  BC  and  bisecting  AB,  it  can  be  dem- 
onstrated that  ABG,  or  its  equal,  DBC,  is  greater  than 
CAB. 

Cor,  Any  two  angles  of  a  triangle  are  together  less 
than  two  right  angles.  Because  the  exterior  angle  is 
greater  than  either  of  the  interior  and  opposite  angles ; 
to  this  inequality  add  the  adjacent  angle,  and  the  sum 
of  the  exterior  angle  and  the  adjacent  angle  w^ill  be 
greater  than  the  sum  of  the  adjacent  angle,  and  either 
of  the  other  two.  But  the  sum  of  the  adjacent  angle  and 
tJie  exterior  angle  is  equal  to  two  right  angles  (Prop. 
XV.).  Therefore  the  sum  of  the  two  angles  of  the  tri- 
angle is  less  than  two  right  angles. 

PROPOSITION  XIX. — Theorem. 

If  two  angles  of  a  triangle  he  equal^  the  sides  opposite 
to  them  are  also  equal. 

Let  ABC  be  a  triangle  having  the 
angle  BAC  equal  to  the  angle  ABC  ; 
then  will  the  side  BC  be  equal  to  the 
side  AC. 

For,  suppose  AC  and  BC  are  not 
equal,  then  one  of  them  must  be  the 
greater;  let  BC  be  the  greater;  then  A  B 

from  BC  (Prop.  III.)  cut  off  a  part,  BD,  equal  to  the 
less,  AC.  In  the  two  triangles  BAC  and  ABD,  the  side 
AC  is  equal  to  BP  (const.),  the  side  AB  common,  and  the 


82  ELEMENTS   OP   PLANE    GEOMETRY. 

angles  BAG  and  ABD  are  equal  (hypothesis.)  There- 
fore the  two  triangles  have  two  sides  and  the  included 
angle  of  the  one  equal  to  two  sides  and  the  included 
angle  of  the  other,  each  to  each.  Hence  (Prop.  I Y.) 
they  are  equal  in  all  their  parts,  which  is  impossible  (Ax. 
9).  Therefore  BD  is  not  equal  to  AC ;  and  in  like  man- 
ner it  can  be  proved  that  no  other  line  than  BC  is  equal 
^o  AC. 

Proposition  XX. — Theorem. 

The  greater  side  of  every  triangle  has  the  greater  angle 
opposite  to  it. 

Let  ABC  be  a  triangle  having  the 

side  AC  greater  than  the   side  BC; 

then  will  the  angle  ABC  be  greater 

than  the  angle  CAB. 

From  the  greater  line,  AC,  cut  off 

CD  equal  to  CB  (Prop.  III).  Join 
the  points  D  and  B.  Because  the  triangle  DCB  is  isos- 
celes, the  angle  CDB  is  equal  to  the  angle  DBC  (Prop. 
VI.).  But  the  angle  CDB  is  greater  than  the  angle  A 
(Prop.  XVni.),  and  hence  the  angle  DBC  must  also 
be  greater  than  the  angle  A.  But  the  angle  ABC  is 
greater  than  DBC  (Ax.  9),  therefore  ABC  is  much  greater 
than  A.  v,        >/. 

Proposition  XXI. — Theorem. 

The  greater  a7igle  of  every  triangle  has  the  greater  side 
opposite  to  it. 

Let  ABC  be  a  triangle  having  the 
angle  ABC  greater  than  the  angle 
BAC;  then  will  the  side  AC  be 
greater  than  the  side  BC. 

For  if  AC, be  not  greater  than  BC,  it 
must  be  equal  to,  or  less  than  it.  It 
can  not  be  equal  to  it ;  for  then  the  angle  ABC  would 
be  equal  to  the  angle  BAC  (Prop.  VI.),  which  is  con- 


BOOK  I.  33 

trary  to  the  hypothesis.  It  can  not  be  less,  for  then 
(Prop.  XX.)  ABC  would  be  less  than  BAG,  which  is  also 
contrary  to  the  hypothesis.  Since  AC  is  neither  equal 
to,  nor  less  than  BC,  it  must  be  greater. 

Propositiox  XXII. — Theorem. 

If ^  from  a  point  icithhi  a  triangle^  two  straight  lines  be 
draicn  to  the  extremities  of  one  side,  these  tico  lines  icill  be 
less  than  the  other  two  sides  of  the  triangle. 

Let  ABC  be  a  triangle,  and  the 
two  lines  AE  and  EB  drawn  from 
the  point  E  to  the  extremities  A  and 
B;  then  will  the  sum  of  AE  and  EB 
be  less  than  the  sum  of  AC  and  CB. 

Produce  AE  to  D.      In  the  trian-      

gle  BDE  the  side  BE  is  less  than   ^  B 

the  sum  of  BD  and  DE  (Prop.  XII.).  To  this  in- 
equality  add  AE ;  then  the  sum  of  BE  ajid  AE  (Ax.  4) 
is  less  than  the  sum  of  BD,  DE,  and  AE,  or  BD  and 
DA.  In  the  triangle  ACD,  the  side  AD  is  les^  than 
the  sum  of  AC  and  CD  (Prop.  XII.).  To  this,  in- 
equality add  BD ;  then  the  sum  of  AD  and  DB  is  less 
than  the  sum  of  AC,  CD,  and  DB,  or  than  the  sum  of 
AC  and  CB.  But  it  was  before  proved  that  the  sum  of 
AE  and  EB  is  less  than  the  sum  of  AD  and  DB;  hence 
the  sum  of  AE  and  EB  is  much  less  than  the  sum  of  AC 
and  CB. 

Gor,  The  two  lines  drawn  from  the  point  E  to  the  ex- 
tremities of  the  side  AB  will  contain  a  greater  angle 
than  the  other  two  sides  of  the  triangle. 

For  the  angle  AEB  is  greater  than  the  angle  EDB ; 
and  EDB  is  greater  than  the  angle  ACD  (Prop.  XYIIL). 
Hence  AEB  is  much  greater  than  ACB. 

B2 


34 


ELEMENTS  OF  PLANE  GEOMETRY. 


PjEioposiTiON  XXIII. — Theorem. 

If  two  triangles  have  two  sides  of  the  one  equal  to  tioo 
sides  of  the  otJier^  each  to  each^  and  the  included  angle  of 
the  one  greater  than  the  included  angle  of  the  other ^  that 
triangle  having  the  greater  included  angle  will  have  the 
greater  side  opposite  to  it, 

C  ©Let  ABC   and  DEF 

be  two  triangles  having 
the  sides  AC  and  CB 
equal  to  ED  and  DF, 
each  to  each,  and  the 
included  angle  ACB 
greater  than  the  in- 
cluded angle  EDF,  then 
will  the  side  AB  be  greater  than  the  side  EF. 

At  the  point  D,  with  the  line  DF,  construct  an  angle, 
FDG,  equal  to  the  angle  ACB  (Prop.  XIV.) ;  make  GD 
equal  to  ED ;  join  the  points  G  and  F,  and  G  and  E.  , 

In  the  two  triangles  ACB  and  GDF,  the  sides  AC  and 
CB  are  equal  to  GD  and  T)F,  each  to  each,  and  the  in- 
cluded angle  ACB  is  equal  to  the  included  angle  GDF 
(const.) ;  therefore  AB  is  equal  to  GF  (Prop.  IV.).  The 
triangle  GDE  is  isosceles,  because  DG  is  equal  to  DE 
(const.) ;  hence  the  angle  DGE  is  eqiial  to  the  angle 
DEG;  but  the  angle  GEF  is  greater  than  the  angle 
GED  (Ax.  9),  and  also  greater  than  DGE;  but  DGE  is 
greater  than  EGF;  hence  GE!F  is  much  greater  than 
EGF;  and,  since  GEF  is  greater  than  EGF,lhe  side  GF 
must  be  greater  than  the  side  EF^.but  GFis  equal  to 
AB.     Therefore  AB  is  greater  than  EF. 


BOOK   I.  35 

Peoposition  XXIV. — Theorem. 

If  tico  triangles  have  tico  sides  of  the  one  equal  to  two 
sides  of  the  other,  each  to  each,  and  the  third  side  of  the 
07ie  greater  than  the  third  side  of  the  other,  that  triangle 
having  the  greater  third  side  icill  have  the  greater  angle 
opposite  to  it. 

Let    the    two    triangles  AC 

ABC  and  DEF  have  two 
sides,  AC  and  CB,  of  the 
one  equal  to  two  sides,  DF 
and  FE,  of  the  other,  each 
to  each,  and  let  the  third 
side,  AB,  be  greater  than 
the  third  side,  DE ;  then  will  the  angle  ACB  opposite 
the  greater  side  be  greater  than  the  angle  DFE  opposite 
the  less  side. 

For,  if  the  angle  ACB  be  not  greater  than  the  angle 
DFE,  it  must  be  equal  to,  or  less  than,  it.  The  angle 
ACB  can  not  be  equal  to  the  angle  DFE,  because  then 
the  side  AB  would  be  equal  tp  the  side  DE  (Prop.  IV.), 
which  is  contrary  to  hypothesis.  It  can  not  be  less,  for 
then  the  side  AB  would  b(j  less  than  DE  (Prop.  XXIII.), 
which  is  also  contrary  to  hypothesis.  Then,  since  the 
angle  ACB  is  neither  equal  to  nor  less  than  the  angle 
DFE,  it  must  be  greater. 

r    Proposition  XXV. — Theorem. 

If  a  straight  line  intersect  two  other  straight  lines,  a7id 
make  the  alternate^  angles  equal,  these  two  straight  lines 
will  he  parallel. 

Let  the  straight  line,  AB,  intersect  the  two  straight 

*  Alternate  literally  means  by  turns ;  in 
Geometry  alternate  angles  are  the  internal 
angles  made  by  two  lines  with  a  third,  on 
opposite  sides  of  it.  A,  A  are  alternate 
angles;  so  are  a,  a.  E,  E  are  exterior 
angles ;  so  are  e,  e. 


36  ELEMENTS  OF  PLANE  GEOMETRY. 

;g  lines,  CD  and  EF,  and  make 

y  the   alternate   angles  EKII 

^ -^ 2^..__^      and  DHK  equal;  then  will 

- — x/_ . ----^^^^^^   these  two  straight  lines  CD 

^y/^  ^  and  EF  be  parallel. 

For,  if  they  are  not  par- 
allel, they  must  meet  if  produced.  Let  them  meet  at- 
any  point,  G.  Then  HGK  becomes  a  triangle ;  and  the 
exterior  angle  (Prop.  XYIII.)  EKH  is  greater  than  the 
interior  and  opposite  angle  KHD;  but  this  is  impos- 
sible; for  they  are  equal  by  hypothesis.  Therefore  CD 
and  EF  can  not  meet  towards  D  or  F,  and  in  the  same 
manner  it  can  be  demonstrated  that  they  can  not  meet 
towards  C  or  E;  and,  since  they  can  not  meet  either 
way,  they  must  be  parallel. 

Peoposition  XXVI. — Theoeem. 

If  one  straight  U7ie  intersect  two  other  straight  lines,  and 
make  the  exterior  angle  equal  to  the  interior  and  opposite 
angle  upon  the  same  side  of  the  line,  or  make  the  sum  of 
the  interior  angles  on  the  same  side  equal  to  tico  right 
angles,  the  tv^o  straight  lines  are  parallel  to  one  another. 

Let  the  straight  line  FE  inter- 

_B    sect  the  two  straight  lines  AB 

and  CD,  and  make  the  exterior 

angle  AHF  equal  to  the  interior 

^  and  opposite  angle,  CGH,  on  til e 
same  side  of  the  line  FE,  or  make 
the  sum  of  the  interior  angles,  AHG  and  CGH,  equal  to 
two  riojht  anscles:  then  the  two  straio^ht  lines  AB  and 
CD  are  parallel. 

For  the  angle  CGH  is  equal  to  the  angle  AHF  (hyp.), 
and  the  angle  BHG  is  also  equal  to  AHF  (Prop.  XYH.) ; 
hence  the  angles  CGH  and  BHG  are  equal  to  each  other 
(Ax.  1);  but  these  are  alternate  angles.  Therefore, 
the  two  straight  lines  AB  and  CD  are  parallel  (Proj). 


BOOK    I.  37 

XXV.).v  Again,  the  sum  of  the  angles  CGII  and  AHG 
is  equal  to  two  right  angles  (hyp.),  and  the  sum  of  the 
angles  AUG  and  BUG  is  also  equal  to  two  right  an- 
gles (Prop.  XV.).  Take  away  the  common  angle  AHG, 
and  there  remain  the  equal  angles  CGH  and  BHG;  but 
these  are  alternate  angles.  Therefore  AB  is  parallel  to 
CD  (Prop.  XXV.). 

Pkoposition  XXYII. — Tiieore^i. 

If  a  straight  line  intersect  two  parallel  straight  lines ^  it 
makes  the  alternate  angles  equal;  it  makes  the  exterior  an- 
gle equal  to  the  interior  and  opposite  ayigle  on  the  same 
sicle^  and  the  sum  of  the  two  interior  angles  on  the  same 
side  equal  to  two  right  angles. 

Let  the  straight  line  Gil  in- 
tersect the  two  parallel  straight 
lines,  AB  and  CD ;  then  will 
the  alternate  angles  CKL  and 
BLK  be  equal :  the  exterior  an- 
gle CKG  will  be  equal  to  the 
interior  and  opposite  angle  ALK  upon  the  same  side,  and 
the  sum  of  the  two  interior  angles  CKL  and  ALK  upon 
the  same  side  be  equal  to  two  right  angles. 

For  if  the  angle  CKL  be  not  equal  to  BLK,  make,  at 
the  point  K,  an  angle,  EKL,  equal  to  it,  and  produce  EK 
to  F ;  and,  since  the  alternate  angles  EKL  and  BLK  are 
equal,  the  lines  EF  and  AB  are  parallel ;  but  (by  hypoth- 
esis) CD  is  parallel  to  AB,  which  is  impossible  (Ax.  12), 
for  two  straight  lines  can  not  be  drawn  through  the  same 
point  parallel  to  the  same  straight  line.  Therefore  CKL 
must  be  equal  to  BLK,  and  ALK  to  DKL.  !N'ow  DKL 
is  equal  to  CKG  (Prop.  XVIL);  hence  ALK  is  also 
equal  to  CKG:  to  each  of  these  equals  add  the  angle 
CKL,  and  the  sum  of  the  angles  ALK  and  CKL  will  be 
equal  to  the  sum  of  the  angles  CKG  and  CKL ;  but  the 
sum  of  CKG  and  CKL  is  equal  to  two  right  angles  (Prop. 


38  ELEMENTS  OP  PLANE  GEOMETRY. 

XV.) ;  therefore  the  sum  of  the  angles  ALK  and  CKL  is 
also  equal  to  two  right  angles. 

Cor,  1.  If  a  straight  line  intersect  two  other  straight 
lines,  and  make  the  sum  of  the  two  interior  angles  on  the 
same  side  less  than  two  right  angles,  these  two  straight 
lines  will  meet  on  the  side  of  the  secant  line  on  which  the 
sum  of  the  two  angles  is  less  than  two  right  angles.  For 
if  they  did  not  meet  when  produced,  they  would  be  par- 
allel ;  and  if  parallel,  the  sum  of  the  two  interior  angles 
on  the  same  side  Avould  be  equal  to  two  right  angles, 
which  is  contrary  to  hypothesis. 

Cor,  2.  If  a  line  be  perpendicular  to  one  of  two  paral- 
lels, it  must  be  perpendicular  to  the  other.  For  if  one  of 
the  alternate  angles  be  a  right  angle,  the  other  must  be  a 
right  angle  also. 

Cor,  3.  When  a  straight  line  intersects  two  parallel 
straight  lines,  all  the  acute  angles  will  be  equal,  and 
also  all  the  obtuse  angles ;  and  the  sum  of  any  acute 
and  obtuse  angle  will  be  equal  to  two  right  angles. 

Peoposition  XXYIII. — Theorem. 
Two  straight  lines  which  are  parallel  to  the  same  straight 
line  are  parallel  to  each  other. 

Let  AB  and  CD  be  parallel 
to  the  same  straight  line  EF; 
then  will  they  be  parallel  to 
each  other. 

Because  the  parallel  lines 
AB  and  EF  are  cut  by  GH, 
the  angle  AJVIL  is  equal  to 
MLF  (Prop.  XXYII);  but 
MLF  is  equal  to  ELK  (Prop.  XVII.) ;  and  since  the  par- 
allel lines  CD  and  EF  are  cut  by  GH,  the  angles  ELK 
and  DKL  are  equal  (Prop.  XXVIL).  Therefore  DKL 
must  be  equal  to  AML;  but  these  are  alternate  angles. 
Hence  AB  and  CD  are  parallel  (Prop.  XXV.). 


BOOK    I.  39 


Proposition  XXIX. — Problem. 

Through  a  given  point  to  draw  a  straight  line  parallel 
to  a  given  straight  line. 

Let  A  be  a  given  point,  and  BC 
a  given  line ;  it  is  required  to  draw 
a  line  tlirough  A  parallel  to  BC. 

Take  any  point,  F,  in  the  line   ^— 

BC,  and  connect  it  with  the  given 


point ;  then  at  the  point  A,  with  the  line  AF,  make  an 
angle,  DAF,  equal  to  the  angle  AFB  (Prop.  XIV.). 
Produce  DA  to  E.  Then  will  the  straight  line  DE  be 
parallel  to  BC. 

For,  since  the  alternate  angles  AFB  and  DAF  are 
equal,  the  lines  DE  and  BC  are  parallel  (Prop.  XXV.). 

Proposition  XXX. — ^T^iieorem. 

If  one  side  of  a  triangle  he  produced^  the  exterior  angle 
is  equal  to  the  sum  of  the  two  interior  and  opposite  angles, 
and  the  three  angles  of  the  triangle  are  together  equal  to 
two  right  angles. 

Let  ABC  be  a  triangle, 
and  let  one  of  its  sides, 
AB,  be  produced  to  D; 
then  will  the  angle  CBD 
bp  equal  to  the  sum  of 
the  interior  and  opposite 
angles  A  and  C ;  and  the 
sum  of  the  three  angles  A,  C,  and  ABC  will  be  equal  to 
two  ris^ht  anojles. 

Through  the  point  B  draw  the  line  BE  parallel  to  AC 
(Prop.  XXIX.) ;  and  because  the  lines  AC  and  BE  are  par- 
allel, and  the  line  BC  intersects  them,  the  alternate  angle 
CBE  is  equal  to  the  alternate  angle  ACB,  and  because 
AD  cuts  them,  the  exterior  angle  DBE  is  equal  to  the  in- 
terior and  opposite  angle,  BAC.     Hence  the  sum  of  the 


40  ELEMENTS    OF   PLANE    GEOMETRY, 

angles  CBE  and  DBE,  or  CBD,  is  equal  to  the  sum  of 
the'  angles  ACB  and  CAB  (Prop.  XXYIL).  Again,  the 
angle  CBD  has  just  been  proved  equal  to  the  sum  of 
the  angles  BAC  and  ACB;  to  each  add  the  angle  ABC, 
and  the  sum  of  the  angles  CBD  and  CBA  is  equal  to 
the  sum  of  "the  angles  BCA,  BAC,  and  ABC;  but  the 
sum  of  the  angles  CBD  and  CBA  is  equal  to  two  right 
angles  (Prop.  XV.) ;  therefore  the  sum  of  the  angles 
ACB,  BAC,  and  ABC  is  equal  to  two  right  angles. 

Cor,  1.  Since  the  three  angles  of  every  triangle  are  to- 
gether equal  to  two  right  angles,  it  follows  that  if  two 
triangles  have  two  angles  of  the  one  equal  to  two  angles 
of  the  other,  the  remaining  angles  must  be  equal. 

Cor,  2.  The  sum  of  all  the  interior  angles  of  a  polygon 
is  equal  to  twice  as  many  right  angles  as  the  figure  has 
sides,  minus  four  right  angles. 

In  the  case  of  the  triangle,  this  corollary  has  just  been 
demonstrated ;  for,  two  right  angles  taken  three  times 
equal  six  right  angles,  from  which  subtract  four  right  an- 
gles, and  two  right  angles  remain. 

In  the  case  of  a  quadrilateral,  draw  a  diagonal,  dividing 
it  into  two  triangles;  it  is  evident  that  the  sum  of  the 
four  angles  will  be  equal  to  four  right  angles — that  is, 
two  right  angles  taken  four  times,  which  make  eight 
right  angles ;  from  which  subtract  four  right  angles,  and 
four  right  angles  remain. 

In  the  case  of  a  pentagon,  draw  two  diagonals,  dividing 
it  into  three  triangles.     The  sum  of  the  interior  angles 

of  the  pentagon  is  equal  to  the 
sum  of  the  angles  of  the  three 
triangles.  But  the  sum  of  the 
angles  of  the  three  triangles 
is  equal  to  six  right  angles 
(Prop.  XXX.) ;  hence  the  sum 
of  the  interior  angles  of  the  pentagon  is  equal  to  six  right 
angles — that  is,  twice  as  many  right  angles  as  the  figure . 
has  sides,  minus  four  right  angles.  And  in  the  same  man- 
ner it  mny  be  demonstrated  of  any  polygon. 


BOOK  I.  41 

Cor.  3.  All  the  exterior  angles  of  any  polygon  are  to- 
gether equal  to  four  right  angles.  It  has  just  been  proved 
that  the  sum  of  all  the  interior  angles  of 
a  pentagon  is  equal  to  six  right  angles ; 
but  the  sum  of  each  interior  and  exterior 
angle  (Prop.  XIV.)  is  equal  to  two  right 
angles ;  therefore  the  sum  of  the  five  in- 
terior and  exterior  angles  must  be  equal 
to  ten  right  angles ;  from  these  ten  right 
angles  subtract  the  six  right  angles  (to 
which  the  sum  of  the  interior  angles  is  equal),  and  there 
remain  four  right  angles,  to  which  the  sum  of  all  the 
exterior  angles  must  be  equal. 

Cor,  4.  Two  angles  of  a  triangle  being  given,  the  third 
may  be  found  by  subtracting  their  sum  from  two  right 
angles. 

Cor,  5.  In  any  triangle  there  can  be  but  one  right 
angle ;  for  if  there  were  two,  the  third  angle  would  be 
nothing. 

Cor,  6.  In  every  right-angled  triangle  the  sum  of  the 
acute  angles  is  equal  to  one  right  angle. 

Cor,  7.  Every  eouilateral  triangle  must  be  also  equi- 
angular (Prop.  VI3(f) ;  and  each  angle  will  be  one-third 
of  two  right  angles,  or  two-thirds  of  one  right  angle. 

Cor,  8.  Since  the  sum  of  the  angles  of  a  quadrilateral 
is  equal  to  four  right  angles,  if  the  angles  are  all  equal 
each  must  be  a  right  angle. 

Proposition  XXXI. — Theorem. 

A  perpendicular  is  the  shortest  line  that  can  he  drawn 
from  a  given  point  to  a  given  line;  two  oblique  linss 
drawn  from  this  point  to  two  points  on  the  line  equidis- 
tant from  the  perpendicular  are  equal^  and  two  oblique 
lines  to  points  unequally  distant  from  the  perpendicular 
are  unequal^  and  the  longer  oblique  Ime  is  farther  from 
the  foot  of  the  perpendicidar, 

.  Let  AB  be  a  given  line  and  C  a  given  point ;  then  Tvill 


42 


ELEMENTS    OF   PLANE    GEOMETRY. 


the  perpendicular  CD  be  the 
shortest  line  that  can  be  drawn 
to  AB ;  and  let  the  oblique  lines 
CE  and  CF  be  equidistant  from 
the  point  D;  then  will  CE  and 
CF  be  equal;  and  also  let  CE 
and  CG  be  unequally  distant 
from  D ;  then  will  CG  be  great- 
er than  CE. 

In  the  triangle  CDE,  the  angle  EDC  is  a  right  angle 
(Def.  10.),  and  therefore  greater  than  the  angle  CED 
(Prop.  XXX.).  But  the  greater  angle  has  the  greater 
side  opposite  to  it  (Prop.  XXI.) ;  hence  CE  is  greater 
than  CD.  In  the  two  triangles  CDE  and  CDF,  the 
sides  ED  and  DF  are  equal  (const.),  the  side  CD  is  com- 
mon, and  the  angle  EDC  is  equal  to  the  angle  FDC 
(Def.  10.).  Hence  the  two  triangles  have  two  sides  and 
the  included  angle  of  the  one  equal  to  two  sides  and 
the  included  angle  of  the  other,  each  to  each,  and  are 
equal  in  all  their  parts  (Prop.  IV.).  Therefore  CE  is 
equal  to  CF.  The  angle  CED  is  an  acute  angle ;  there- 
fore CEG  is  obtuse  (Prop.  XV. ) ;  and  if  CEG  is  obtuse, 
CGE  must  be  acute  (Prop.  XXX.).  Hence  CG,  opposite 
the  greater  angle,  is  greater  than  CE,  opposite  the  less 
(Prop.  XXL). 


Proposition  XXXII. — ^Theorem. 

If  two  right-angled  triangles  have  the  hypothenuse  and 
a  side  of  the  one  equal  to  the  hypothenuse  and  a  side  of 
the  other ^  each  to  each^  the  two  right-angled  tria?igles  are 
equal  in  all  their  parts. 

Let  the  two  right-angled  triangles  ABE  and  CDF 
have  the  hypothenuse  AE  and  the  side  EB  of  the  one 
equal  to  the  hypothenuse  CF  and  the  side  FD  of  the 
other;   then  will  the  two  triangles  AEB  and  CFD  be 

equal. 


BOOK    I. 


43 


If  AB  were  equal  to  CD  the 
equality  of  the  triangles  would 
be  manifest.  Now  suppose  CD 
not  equal  to  BA,  and  let  CD 
be  the  greater ;  then  from  CD 
cut  off  a  part,  GD,  equal  to 
AB  (Prop.  III.).  Join  G  and 
F.  In  the  two  triangles  ABE 
and  GDF,  the  sides  BE  and  FD  are  equal  (hyp.),  andBA 
and  GD  are  also  equal  (const.) ;  and  the  included  angles 
B  and  D  are  equal,  being  right  angles.  Plence  the  two 
triangles  are  equal  in  all  their  parts  (Prop.  IV.) ;  GF  is 
equal  to  AE ;  but  FC  is  equal  to  AE  (hyp.) ;  therefore 
FC  is  equal  to  FG ;  but  this  is  impossible  (Prop.  XXXI.). 
Hence  GD  can  not  be  equal  to  AB;  and  in  the  same 
way  it  can  be  proved  that  no  other  line  except  CD  can ' 
be  equal  to  AB.  The  two  triangles  ABE  and  CDF  are, 
therefore,  every  way  equal. 


Peoposition  XXXIII. — Theorem. 

If  tioo  angles  have  their  sides  parallel^  each  to  each^  and 
lying  in  the  same  direction^  they  are  equal. 

If  the  straight  lines 
AC  and  CB  be  parallel 
to  ED  and  EF,  each  to 
each,  and  lie  in  the  same 
direction,  then  will  the 
angle  ACB  be  equal  to 
the  angle  DEF. 

For,  draw  the  line  ECG  through  their  vertices ;  and 
since  AC  is  parallel  to  DE,  the  exterior  angle  ACG  is 
equal  to  the  interior  and  opposite  angle,  DEC,  on  the 
same  side  (Prop,  XXVII.) ;  and  since  BC  is  parallel  to 
FE,  the  exterior  angle  BCG  is  also  equal  to  the  interior 
and  opposite  angle,  FEC,on  the  same  side  (Prop.  XXVII.). 
If  from  the  equal  angles  ACG  and  DEC,  the  equal  angles 


44  ELEMENTS    OF   PLANE    GEOMETRY. 

BCG  and  FEC  be  subtracted,  there  remain  AGB  and  DEF 
equal. 

Cor,  If  AC  and  BC  be  produced  to  H  and  I,  the  angle 
HCI  will  be  equal  to  ACB ;  but  DEF  is  also  equal  to 
ACB.     Hence  HCI  and  DEF  are  equal. 

Scholium,  The  restriction  that  the  parallel  sides  must 
lie  in  the  same  direction  is  necessary ;  for  the  angle  BCH 
has  its  sides  parallel  to  the  sides  of  the  angle  DEF,  but  is 
not  equal  to  it. 


The  following  test  examples^  involving  the  First  Boole 
only^  are  solved  with  the  view  of  giving  the  learner  an  in- 
sight into  the  methods  of  geometrical  application  of  pre- 
vioics  2^Tinciples, 

1.  To  trisect  a  giveJi  line,"^ 

Let  AB  be  the  given  line;  it  is 

required  to  trisect  it. 

On  AB  construct  an  equilateral 

triangle,  ABC  (Prop.  I.) ;  bisect  the 

angles  BAC  and  ABC  (Prop.  VIII.) ; 

from  the  point  of  intersection,  D, 

draw  DE  and  DF  respectively  par- 

A       E  F        B    alleltoACandCB.    Then  will  these 

lines  trisect  AB  in  the  points  E  and  F.  For  the  angle 
DAE  is  equal  to  CAD  (const.),  and  ADE  is  also  equal 
to  CAD  (Prop.  XXVII. ).  Therefore,  since  each  is  equal 
to  the  same,  they  are  equal  to  each  other ;  DAE  equals 
ADE.  Hence  the  side  AE  equals  the  side  ED  (Prop. 
XIX.),  and  similarly  FB  can  be  proved  equal  to  DF. 
The  angles  DEF  and  DFE  are  equal  respectively  to  the 
angles  CAB  and  ABC  (Prop.  XXVIL).  Therefore  the 
angle  EDF  is  equal  to  the  angle  C.  But  the  triangle 
ABC  is  equilateral  and  equiangular.     Hence  the  triangle 

*  This  problem  is  usually  and  easily  solved  by  the  subsequent  proposi- 
tion, that  the  lines  parallel  to  the  base  divide  the  other  sides  proportion- 
ally ;  but  we  are  limited  to  the  propositions  of  the  First  Book. 


BOOK    I.  45 

EDF  is  also  equilateral  and  equiangular.  Therefore  EF 
is  equal  to  ED  and  DF,  or  to  their  equals,  AE  and  FB. 
Hence  AB  is  trisected. 

2.  Through  a  given  'point  to  draw  a  straight  tine  which 
shall  make  equal  angles  tcith  two  straight  lines  given  in 
positio7i. 

Let  E  be  the  given 
point,  and  AB  and  CD 
the  two  lines  given  in 
position. 

Prolong  AB  ^nd  CD 
until  they  meet  in  the 
point  F;  bisect  the  an- 
gle BFD;*    from  the   point  E  draw  the  perpendicular 
GE,  and  produce  it  both  ways  to  B  and  D ;  then  will 
the  line  DB  make  equal  angles  with  the  lines  AB  and  CD. 

For  in  the  two  triangles  GBF  and  GDF,  the  angles 
GFB  and  GFD  are  equal  (const.),  and  the  angles  BGF 
and  DGF  are  equal,  being  right  angles.  Therefore  the 
remaining  angles,  GBF  and  GDF,  are  equal. 

3.  From  two  given  points  on  the  same  side  of  a  line 
given  in  position  to  draw  two  lines  which  shall  meet  in 
that  liiie  and  make  equal  angles  with  it. 

Let  A  and  B  be  the 
given  points,  and  CK 
the  given  line  in  posi- 
tion. 

From  the  point  A  let 

fall    the     perpendicular 

^  j  AE ;   prolong  AE  until 

ED  is  equal  to  it.  Join  D  and  B,  and  A  and  F.  Then 
the  required  angles  are  AFE  and  BFK. 

*  It  is  supposed  that  the  student  is  sufficiently  familiar  with  the  prob- 
lems required  in  this  solution  without  naming  them. 


D 


46  ELEMENTS    OF   PLANE    GEOMETRY. 

For,  in  the  two  triangles  AFE  and  DFE,  AE  is  equal 
to  ED  (const.),  EF  is  common,  and  the  included  angles 
AEF  and  DEF  are  equal,  being  right  angles.  Hence 
the  angle  AFE  is  equal  to  the  angle  DFE;  but  the  an- 
gle DFE  is  equal  to  the  angle  BFK.  Therefore  AFE 
is  also  equal  to  BFK. 


TEST  EXAMPLES  IN  BOOK  I. 

1.  Given  two  angles  of  a  triangle,  to  find  the  third 
angle. 

2.  To  construct  an  isosceles  triangle  with  a  given  base 
and  vertical  angle. 

3.  To  trisect  a  right  angle — that  is,  to  divide  it  into 
three  equal  parts. 

4.  Prove  that  every  point  of  the  bisectrix  of  an  angle 
is  equally  distant  from  the  sides. 

5.  The  three  straight  lines  drawn  from  a  point  within 
a  triangle  to  the  angles  are  together  less  than  the  perim- 
eter, but  greater  than  its  half. 

6.  To  construct  an  isosceles  triangle,  so  that  the  vertex 
w^ill  fall  at  a  given  point  and  the  base  fall  in  a  given  line. 

Y.  Given  the  perpendicular  of  an  equilateral  triangle,  to 
construct  it. 

8.  Given  the  diagonal  of  a  square,  to  construct  it. 

9.  To  construct  an  isosceles  triangle,  so  that  the  base 
shall  be  a  given  line,  and  the  vertical  angle  a  right  angle. 

10.  Given  the  sum  of  the  diagonal,  and  a  side  of  a 
square,  to  construct  it. 

11.  To  construct  a  triangle  when  the  altitude,  the  ver- 
tical angle,  and  one  of  the  sideSy  are  given. 

12.  Given  the  sum  of  the  three  sides  of  a  triangle,  and 
the  angles  at  the  base,  to  construct  it. 

13.  From  two  given  points  to  draw  two  equal  straight 
lines  which  shall  meet  in  the  same  point  of  a  line  given 
in  position.  

A  KEY  TO  THE  TEST  EXAMPLES  IN  BOOK  L 

1.  See  Prop.  XXX. 

2.  See  Props.  XXX.,  XIV.,  and  XV. 

3.  See  ProT)5. 1,  and  VIIT. 


BOOK    I. 


47 


4.  Sec  Props.  :^XXi:  and  V. 

5.  See  Props.  XXIT.  and  XII. 

6.  See  Props.  XL,  IX.,  and  IV. 

I.  See  Test  Ex.  Ill,  and  Prop.  X. 

8.  See  Prop.  VIII.     Bisect  right  angles. 

9.  Bisect  right  angle  Prop.  ^11.,  and  see  Prop.  XIV. 

10.  See  the  accompanying 
figure :  AB  equals  sum  of 
diagonal  and  side.  At  B 
make  an  angle  equal  to  one- 
quarter  of  a  right  angle.  At 
A  make  BAE  equal  to  one- 
half  a  riglit  angle.  The  stu- 
dent will  readily  solve  the  re- 
mainder. 

II.  First  take  a  =  al- 
titude, s  given  side, 
and  /v_  the  given  angle ;  then 
draw  a  base  line  of  any  length,  and  by  Prop.  X.  erect  per- 
pendicular =  to  a.  With  the  upper  extremity  as  centre, 
and  the  length  of  s  as  radius,  describe  an  arc  cutting  the 
base.  Then,  by  means  of  Props.  XXXI.  and  XIV.,  the 
solution  becomes  simple. 

12.  See  Props. 
VIIL,XIV.,XIX., 
and  XXX. 


13.  See  Props.  IX.  and  X.,  and  the 
accompanying  figure ; 


4a 


ELEMENTS    OF   PLANE    GEOMETRY. 


BOOK  IL 


DEFINITIONS. 

A  RIGHT-ANGLED  Parallelogram,  or  Rectangle,  is  said 
to  be  contained  by  any  two  of  the  straight  lines  which 
are  about  one  of  the  right  angles. 

Thus  the  rectangle  ADCB 
is  said  to  be  the  rectangle  con- 
tained by  AD  and  DC,  or  by 
AD  and  AB,  etc. 

For  the  sake  of  brevity,  the 
rectangle  ADCB  is  usually 
called  the  rectangle  AC  or 
DB ;  and  instead  of  the  rectangle  contained  by  AD  and 
DC,  it  is  simpler  to  say  the  rectangle  AD .  DC,  placing  a 
point  between  the  two  sides  of  the  rectangle. 

In  Geometry,  the  product  of  two  lines  means  the  same 
thing  as  their  rectangle. 


D 


Proposition  I. — Theorem. 

Two  straight  lines  which  joi7i  the  extremities  of  two 
equal  a7icl  parallel  straight  lines^  towards  corresponding 
parts ^  are  also  equal  and  parallel. 

Let  AB  and  CD 
be  two  equal  and 
parallel  straight 
lines;  and  let  AC 
and  BD  join  their 
corresponding  ex- 
tremities ;  then  will 

AC  and  BD  be  equal  and  parallel. 

Draw  AD.     And  because  AB  and  CD  are  parallel,  and 

AD  intersects  them,  the  alternate  angles  BAD  and  ADC 


^  BOOK    II.  49 

ire  equal  (Prop.  XXVIL,  Bk.  I.).  Xow,  in  the  two  tri- 
^  angles  BAD  and  ADC,  the  sides  AB  and  CD  are  equal 
(hyp.),  the  side  AD  is  common,  and  the  included  angles 
BAD  and  ADC  are  also  equal.  Hence  the  two  triangles 
are  every  way  equal ;  BD  is  equal  to  AC,  and  the  angle 
CAD  is  equal  to  the  angle  ADB ;  but  these  being  alter- 
nate angles,  AC  and  BD  are  parallel  (Prop.  XXVI., 
Bk.  I). 

Cor,  If  two  sides  of  a  quadrilateral  are  equal  and  par- 
allel, the  figure  will  be  a  parallelogram. 

Pkopositiox  II. — The  DEEM. 

If  the  opposite  sides  of  a  quadrilateral  are  equal^  each 
to  each^  the  figure  is  a  parallelogram. 

Let     the     opposite  q  jy 

sides,  AB  and  CD,  be 
equal,  and  also  AC 
and  BD ;  then  will  the 
quadrilateral  ABDC 
be  a  parallelogram. 

For,  having  drawn 
BC,  the  two  triangles  ABC  and  DCB  have  three  sides 
of  the  one  equal  to  three  sides  of  the  other,  each  to  each ; 
hence  they  are  every  way  equal  (Prop.  VII.,  Bk.  I.) ;  and 
the  angle  ABC  is  equal  to  the  angle  BCD :  these  being 
alternate  angles,  AB  and  CD  are  parallel  (Prop.  XXVJt) ; 
and  the  angle  ACB  is  equal  to  the  angle  CBD ;  and  these 
being  also  alternate  angles,  AC  and  BD  are  parallel 
(Prop.  XXVl;  Bk.  I).  Hence  the  figure  ABDC  is  a 
parallelogram. 

Cor.  If  the  opposite  angles  of  a  quadrilateral  be  equal, 
each  to  each,  the  figure  is  a  parallelogram. 

For  all  the  angles  of  the  figure  being  equal  to  four 
right  angles  (Cor.  4[^  Prop.  XXX.,  Bk.  I.),  and  the  oppo- 
site angles  being  mutually  equal,  each  pair  of  adjacent 
angles  must  be  together  equal  to  two  right  angles;  but 
these  adjacent  angles  become  the  interior  angles  on  the 

C 


50  ELEMENTS  OF  PLANE  GEOMETKY. 

same  side;   hence  the  opposite  sides  must  be  parallel \ 
(Prop.  XXVI.,  Bk.  L). 

Proposition  III. — Theorem. 

The  opposite  sides  of  a  parallelogram  are  equal,  each  to 
each,  and  the  diagonal  bisects  it. 

Let  ABDC  be   a 

parallelogram;  then 
will  the  opposite 
sides  AB  and  CD  be 
equal,  and  also  AC 
and  BD ;  and  the  di- 
agonal CB  will  bi- 
sect the  figure. 
Because  AB  is  par- 
allel to  CD,  and  BC  intersects  them,  the  angle  ABC  is 
equal  to  the  angle  BCD ;  and  because  AC  is  parallel  to 
BD,  and  CB  intersects  them,  the  angle  ACB  is  equal  to 
the  angle  CBD  (Prop.  XXYII.,  Bk.  t).  IsTovv  in  the  two 
triangles  ABC  and  DCB,  the  side  CB  is  common,  and  the 
angles  ACB  and  ABC  are  equal  to  CBD  and  BCD,  each 
to  each.  Hence, the  triangles  are  equal  in  all  their  parts 
(Prop,  v.,  Bk.  I.) — that  is,  the  diagonal  bisects  the  figure, 
and  the  opposite  sides  are  equal. 

Cor.  1.  The  opposite  angles  of  a  parallelogram  are 
equal.  For  the  angle  A  is  equal  to  D,  and  the  sum  of 
ACB  and  BCD  or  ACD  is  equal  to  the  sum  of  CBD  and 
CBA  or  ABD. 

Cor.  2.  Two  parallel  lines  included  between  two  other 
parallel  lines  are  equal. 

Cor.  3.  Hence  two  parallels  are  everywhere  equally 
distant. 

Proposition  IY. — Theorem. 

The  diagonals  of  a  parallelogram  mutually  bisect  each 
other. 

Let  ABDC  be  a  parallelogram ;  then  will  its  diagonals, 
AD  and  CB,  mutually  bisect  each  other  at  the  point  E. 


BOOK   II.  51 

For,  ill  the  two 
triangles  ABE  and 
DCE,  the  sides  AB 

^d  DC   are    equal 

Dyhyp.);  the  angle 
ABE  is  equal  to 
DCE  (Prop.  XXVIL, 
Bk.  I.) ;  and  for  the 
same  reason  the  an-  ^ 
gle  BAE  is  equal  to  the  angle  CDE :  the  two  triangles 
liave,  therefore,  two  angles  and  the  included  side  of  the  one 
equal  to  two  angles  and  the  included  side  of  the  other,  each 
to  each ;  hence  they  are  equal  in  all  their  parts  (Prop.  Y., 
Bk.  I.) ;  the  side  AE  is  equal  to  ED,  and  BE  to  EC. 

Proposition  V. — ^Theorem. 

If  the  diagonals  of  a  quadrilateral  inictually  bisect  'each 
other ^  the  figure  is  a  parallelogram. 

Let  ABDC  be  a  quadrilateral  whose  diagonals  mutu- 
ally bisect  each  other  at  the  point  E ;  then  will  the  iig- 
UHJ  be  a  parallelogram. 

For  in  the  two  triangles  AEB  and  CED,  AE  is  equal 
to  ED,  and  BE  to  EC  (hyp.),  and  the  included  angles 
AEB  and  CED  are  equal  (Prop.  XYIL,  Bk.  I.).  Hence 
they  are  equal  in  all  their  parts  (Prop.  lY.,  Bk.  I.) ;  AB 
is  equal  to  CD,  and  in  the  same  way  it  may  be  demon- 
strated that  AC  is  equal  to  BD.  Therefore  ABDC  is  a 
parallelogram  (Prop.  II.). 

^  Proposition  YI. — Theorem: 

Parallelograms  upon  the  same  base,  and  between  the 
same  parallels^  are  equal,. 

Let  the  two  parallelograms  ABCD  and  ABEF  have 
the  same  base,  AB,  and  be  between  the  same  parallels 
AB  and  FC,  then  will  th^se  two  parallelograms  be  equal. 

In  the  two  triangles  EBC  and  FAD,  the  sides  BC  and 
AD  are  equal,  and  BE  and  AF  are  equal  (Prop.  IIL),  and 


52  ELEMENTS  OF  PLAXE  GEOMETRY. 

the  angles  CBE  and  FAD  are  equal  (Prop.  XXXIII. ,  Bk. 
I.).  Hence  these  triangles  have  two  sides  and  the  in- 
cluded angle  of  the  one  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other,  each  to  each ;  they  are,  there- 

F E       D C  3i'        D       E         C 


A  B  A.  B 

fore,  equal  in  all  their  parts.  If,  now,  from  the  whole 
figure  ABCF  the  triangle  BCE  be  taken,  there  will  re- 
main the  parallelogram  ABEF,  and  if  the  equal  triangle 
AFD  be  taken  from  the  same  figure,  there  will  remain 
the  parallelogram  ABCD.  Hence  the  parallelograms  are^* 
equal  (Ax.  3). 

Cor.  Parallelograms  having  the  same  base  and  the 
same  altitude  are  equal;  for,  having  the  same  altitude, 
or  perpendicular  height,  they  must  be  between  the  same 
parallels  (Cor.  3,  Prop.  HI.). 

Peoposition  VII. — Theorem. 
Para llelocf rams  having  equal  bases,  and  beticeen  the 
same  2^(jirallels,  are  equal, 

Ti  C  TT  G-       I^et  the   parallelo- 

grams ABCD  and 
EFGH  have  equal 
bases,  AB  and  EF, 
and  be  between  the 
same  parallels,  AF 
and  DG;  then  will 
these  parallelograms 


.A.  B       IE  I* 

be  equal. 


BOOK   II.  o3 

Join  the  points  A  and  II,  and  B  and  G.  AB  is  equal 
to  EF  (by  hyp.),  and  HG  is  equal  to  EF  (Prop.  III.) ; 
hence  AB  and  HG  are  equal  (Ax.  1).  HG  and  AB  being 
equal  and  parallel,  the  lines  AH  and  BG,  which  join  thc^ir 
corresponding  extremities,  must  also  be  equal  and  paral- 
lel (Prop.  I.).  Therefore  ABGH  is  a  parallelogram  ;  and 
the  parallelogram  ABCD  is  equal  to  this  parallelogram, 
ABGH  (Prop.  YI.),  and  for  the  same  reason  EFGH  is 
equal  to  ABGH.     Hence  ABCD  and  EFGH  are  equal 

Proposition  VIH. — Theor:i2M.        /^^ 

Triangles  upon  the  same  hase^  and  between  the  same 
parallels^  are  equaL 

Let  the  two  triangles 
ABC  and  ABD  have  the 
same  base,  AB,  and  be 
between  the  same  paral- 
lels, AB  and  CD;  then 
will  these  two  triangles 
be  equal. 

Produce  CD  both  ways, 
and  make  the  line  AF  par- 
allel to  BC  (Prop.  XXIX., 
Bk.  I.) ;  make  BE  also  parallel  to  AD.  The  parallelo- 
grams ABCF  and  ABED  are  equal  (Prop.  VI.) ;  but  the 
triangle  ABC  is  half  the  parallelogram  ABCF,  and  the 
tnangle  ABD  is  half  the  parallelogram  ABED  (Prop. 
HI.) ;  and  because  the  halves  of  equal  magnitudes  are 
equal  (Ax.  7)  the  triangle  ABC  is  equal  to  the  triangle 
ABD. 

Proposition  IX. — Theorem. 

Triangles  upon  equal  bases,  and  between  the  same  par* 
allels,  are  equal. 

Let  the  triangles  ABC  andDEF  stand  upon  equal  bases, 
AB  and  DE,  and  be  between  the  same  parallels,  AE  and 
CF ;  then  will  these  triangles  be  equal. 


54 


ELEMENTS  OF  PLANE  GEOMETRY. 


Through  the  G- 
points  A  and  E 
draw  AG  and  EH 
parallel  to  BC  and 
DF,  each  to  each 
(Prop.XXIX.,Bk. 
I.),  and  produce 
CF  both  ways  to 
G  and  H;  then 
the  figures  ABCG 
and  DEHF  are  parallelograms,  and  are  equal  to  each 
other  (Prop.  VII.),  because  they  are  upon  equal  bases, 
AB  and  DE,  and  between  the  same  parallels,  AE  and 
GH.  But  the  triangle  ABC  is  half  the  parallelogram 
ABCG,  and  the  triangle  DEF  is  half  the  parallelogram 
DEHF  (Prop.  III.),  and  because  the  halves  of  equal  mag- 
nitudes are  equal  (Ax.  1)  the  triangle  ABC  is  equal  to 
the  triangle  DEF. 

Proposition  X. — Theorem. 

If  a  triangle  and  a  parallelogram  he  upon  the  same 
base  and  between  the  same  parallels^  the  triangle  is  half 
the  parallelogram. 

Let  the  triangle  ABC  and 
the  parallelogram  ABDE 
stand  upon  the  same  base, 
AB,  and  be  between  the 
same  parallels,  AB  and  ED ; 
then  will  the  triangle  ABC 
be  half  the  parallelogram 
ABDE. 

Draw  the  diagonal  AD; 
then  the  triangle  ABC  is 
equal  to  the  triangle  ABD  (Prop.  VIII.),  because  they 
are  upon  the  same  base,  AB,  and  between  the  same  par- 
allels, AB  and  ED :  but  the  triangle  ABD  is  half  the 
parallelogram  ABDE  (Prop.  III.) ;  hence  the  triangle 
ABC  is  also  half  the  parallelogram  ABDE. 


ROOK    II. 


55 


Proposition  XL — Theorem. 
TJie  complements^  of  the  parallelograms  which  are  about 
the  diagonal  of  any p)arallelogram  are  equal  to  each  otJier, 

Let  AD  be  a  parallelo- 
gram, BC  its  diagonal, 
FIG  parallel  to  AB  or 
CD,  and  HIE  parallel  to 
AC  orBD;  FE  and  HG 
the  parallelograms  about 
A  H  B  BC,  and  AI  and  ID  the 

complements  of  the  whole  figure  AD;  then  will  AI  be 
equal  to  ID. 

The  triangle  ABC  is  equal  to  the  triangle  CDB,  FCI 
to  lEC,  and  HBI  to  IGB,  because  the  diagonal  bisects 
the  three  parallelograms  AD,  FE,  and  HG  (Prop.  IIL). 
If  from  the  triangle  ABC  the  sum  of  FCI  and  HBI  be 
taken,  and  if  from  the  triangle  CDB  the  sum  of  lEC  and 
IGB  be  taken,  the  complements  AI  and  ID  remain ;  they 
are  therefore  equal.  <>J^^^  ^Of 

Proposition  XII. — Problem.  >*- 

To  co7istruct  a  square  upon  a  given  straight  line. 
C  Let  AB  be  the  given   straight 

j3 1£  line;  it  is  required  to  construct  a 

square  upon  it. 

From  the  point  A  erect  the  per- 
pendicular AC  (Prop.  X.,  Bk.  I.), 
and  make  AD  equal  to  AB,  and 
through  the  point  D  draw  DE  par- 
allel to  AB,  and  through  B  draw 
BE  parallel  to  AD  (Prop.  XXIX., 
Bk.  L).     ABED  is  a  square. 

*  The  student  will  perceive  what  is  meant  by  ** complements"  if  he 
examine  carefully  the  application  of  the  enunciation  to  the  figure.  FE 
and  HG  are  about  the  diagonal,  equal  portions  being  on  either  side  of 
it;  and  AI  and  ID  are  the  "complements, "which,  added  to  the  paral- 
lelograms FE  and  IIG,  make  the  whole  parallelogram  AD. 


B 


56 


ELEMENTS    OF   PLANE    GEOMETRY. 


For,  since  DE  is  parallel  to  AB,  and  BE  to  AD,  ABED 
is  a  parallelogram  i  but  AD  is  equal  to  AB  (const.) ; 
hence  the  four  sides  are  equal  (Prop.  III.),  and  the  figure 
is  equilateral.  The  angle  A  is  a  right  angle;  therefore 
the  other  three  angles  are  right  angles  (Cor.  1,  Prop.  HI.)? 
and  the  figure  is  rectangular.  The  figure  ABED  is  there- 
fore a  square,  and  is  constructed  upon  AB. 

Proposition  XIII. — Theorem. 

.In  a7iy  right-angled  triangle,  the  square  described  upo7i 
the  hypothenuse  is  equal  to  the  sum  of  the  squares  de* 
scribed  upon  the  other  two  sides. 

Let    ABC    be     a  /J^ 

right-angled  triangle, 
and  the  angle  ACB 
the  right  angle ;  then 
will  the  square  of 
AB  be  equal  to  the 
sum  of  the  squares 
of  AC  and  CB. 

Upon  AB,  AC,  and 
CB  describe  the 
squares  AE,  AK,  and 
BH  (Prop.  XII.); 
through  C  draw  CG 
(Prop.  XXIX.,  Bk. 
I.)  parallel  to  AF  or 
BE,  and  join  the 
points  A  and  D,  and  C  and  E. 

The  angles  CBD  and  ABE  are  equal,  because  they  are 
right  angles  (Ax.  11) ;  to  each  add  the  angle  ABC;  and 
the  angle  ABD  will  be  equal  CBE  (Ax.  2).  In  the  two 
triangles  ABD  and  CBE,  the  sides  AB  and  BE  are  equal, 
because  they  are  sides  of  the  same  square ;  and  for  a  simi- 
lar reason  the  sides  BD  and  BC  are  also  equal.  Hence 
these  two  triangles  have  two  sides,  AB  and  BD,  and  the 
included  angle,  ABD,  of  the  one  equal  to  two  sides,  EB 


BOOK   II. 


57 


and  BC,  and  the  included  angle,  EBC,  of  the  other,  each 
to  each ;  they  are,  therefore,  equal  in  all  their  parts  (Prop. 
IV.,  Bk.  I.).  But  the  parallelogram  BG  is  double  the  tri- 
angle EBC,  because  they  are  upon  the  same  base  and  be- 
tween the  same  parallels  (Prop.  X.),  and  the  square  BH 
is  double  the  triangle  ABD  for  a  similar  reason ;  and 
because  the  doubles  of  equal  magnitudes  are  equal  (Ax. 
6),  the  parallelogram  BG  is  equal  to  the  square  BH.  In 
like  manner  it  may  be  demonstrated  that  the  parallelo- 
gram AG  is  equal  to  the  square  AK.  Hence  the  sum 
of  these  parallelograms,  or  the  square  BF,  is  equal  to  the 
sum  of  the  two  squares  BH  and  AK;  but  BF  is  the 
square  of  AB,  AK  of  AC,  and  BH  of  BC.  Therefore  the 
square  of  AB  is  equal  to  the  sum  of  the  squares  of  AC 
and  CB. 


PiiOPosmoN  XIV. — Theorem. 

If  there  he  two  straight  liiies,  one  of  which  is  divided 
into  any  number  of  parts  ^  the  rectangle  contahied  by  the 
two  lines  is  equal  to  the  sum  of  the  rectangles  contained 
by  the  U7idivided  line  and  the  several  parts  of  the  divided 
line. 

Let  A  and  BC  be  two 
straight  lines,  and  let  BC 
y  be  divided  into  any  num- 
ber of  parts  in  the  points 
G  and  H;  the  rectangle 
contained  by  A .  BC  is 
equal  to  the  sum  of  the 
rectangles  A .  BG,  A  .  GH, 
A.HC. 

From  the  point  B  erect 
BD  perpendicular  to  BC 
(Prop.  XL,  Bk.  I.),  make  BE 
equal  to  A  (Prop.  HI.,  Bk.  L),  tlirough  the  point  E  draw 
EF  parallel  to  BC  (Prop.  XXIX.,  Bk.  I.),  make  EF  equal 
to  BC,  and  through  tlie  points  G,  H,  C,  draw  GK,  IlL, 

C2  ■ 


58  ELEMENTS    OF   PLANE    GEOMETRY. 

CF,  parallel  to  BE ;  BF,  BK,  GL,  and  HF  are  rectangles, 
andBF=BK4-GL+HF. 
But  BFrz:BE .  BCz=:A .  BC,  because  BErrA. 
BK=BE.  BG=A .  BG,  for  the  same  reason. 
GL^zGK.  HG=A .  HG,  because  GK=BE=A. 
HF=HL.HC=:A.HC,  because  HL=GK=A. 
HenceA.BC=:A.BG+A.HG+A.HC. 
Scholium,  Propositions  like  the  above  are  easily  de- 
rived from  Alge-  -r  ce "b e  a, 

bra.  Let  the  seg-  __^ 

ments  of  BC  be  denoted  by  a,  ^j  c/  ^ 

then  A  {a-\-h+c)—Aa+Ah+Ac; 

that  is,  A.BC     =:A.a+A.  J+A.c. 


Peoposition  XV. — Theorem. 

If  a  straight  line  he  divided  into  two  parts^  the  square 
of  the  whole  line  is  equal  to  the  sum  of  the  rectangles  con- 
tained by  the  whole  line  and  each  of  the  parts. 

Let  the  straight  line  AB  be  divided  S- 
into  any  two  parts  at  the  point  C ;  the 
square  of  AB  is  equal  to  the  sum  of  the 
rectangles  contained  by  AB .  AC,  and 
AB.CB;  or  AB'=AB.AC+AB.CB. 

On  AB  describe  the  square  AD  (Prop. 
XII.),  and  through  C  draw  CE  parallel  A" 
to  AG  (Prop.  XXIX.,  Bk.  I).  The  square  AD=AE+CD. 
But  ADzr:  AB',  and  AE=AB .  AC,  because  AB=AG ;  and 
CD=: AB .  CB,  because  AB=BD.  Therefore  AB"=AB . 
AC+ AB.CB. 

Scholium,  The  Algebraic  demonstration  is  very  simple : 
Let  «=AB,  h=zAC,  and  c=CB;  then  a=h+c.  Multiply 
both  members  of  the  equation  by  a,  and  we  shall  have  a^ 
z=^ab-\-ac. 


BOOK  II.  b\) 


Proposition  XVI. — Theorem. 

J^  a  straight  line  he  divided  into  two  parts,  the  square 
of  the  whole  line  is  equal  to  the  sum  of  the  squares  of  the 
parts  a7id  twice  the  rectangle  contained  by  the  parts, 

'  Let  AB  be  a  straight  line  divided 
into  two  parts,  AC  and  CB ;  then  will 
the  square  of  AB  be  equal  to  the  sum 
of  the  squares  of  AC  and  CB,  and 
twice  the  rectangle  contained  by  AC 
and  CB.  Or  AB^=AC^+CB'^+2AC 
ir — i  xCB. 
On  AB  describe  the  square  AE  (Prop.  XII.),  on  AC 
describe  the  square  AG,  produce  CG  and  KG  to  F  and 
L.  KL=AB  and  CF=BE  or  AB.  Hence  KL=CF. 
But  KG=GC,  being  sides  of  a  square.  If  from  KL  and 
CF,  KG  and  GC  be  subtracted  respectively,  GL  will  re- 
main =to  GF.  But  GL  and  GF  are  respectively  equal 
to  FE  and  EL.  Hence  GE  is  a  square,  and  since  LG= 
CB,  it  is  the  square  of  CB.  KG=GC  and  GF=GL: 
therefore  the  rectangles  KF  and  CL  are  equal.  But  KG 
and  GC  are  each  equal  to  AC,  and  GF  and  GL  are  each 
equal  to  CB.  Hence  the  rectangles  KF  and  CL  are  twice 
the  rectangle  contained  by  AC  and  CB.  The  whole 
square  AE  is  equal  to  the  squares  AG  and  GE  and  the 
tw^o  rectangles  KF  and  CL.  Therefore  the  square  of 
AB=the  square  of  AC,  the  square  of  BC  and  twice  the 
rectangle  contained  by  AC  and  CB.  Or  AB^=:AG'+ 
CB''+2  AC.CB. 

Algebraically:  Let  AC = a,  and  CB  =  5/  then  (a+&)* 

Cor,  If  the  line  be  divided  into  two  equal  parts,  the 
square  of  the  whole  line  equals  four  times  the  square  of 
half  the  line. 


60  elements  of  plane  geometry. 

Proposition  XVIL — Theorem. 

The  square  of  the  difference  of  two  lines  is  equal  to  the 
sum  of  their  squares  diminished  by  twice  the  rectangle  con- 
tained hy  the  lines. 

Let  AB  and  BC  be  two  lines  whose     q  ^ 

diiFerence  is  AC ;  then  will  the  square 
of  AC  be  equal  to  the  sum  of  the 
squares  of  AB  and  BC  diminished  by 
twice  the  rectangle  of  AB  and  BC :  or 
AC2=AB2+BC2-2AB.  BC. 


5 


On  AC  construct  the  square  AE,  on  A 
AB  construct  the  square  AH,  and  on 
BC  construct  the  square  CL  (Prop. 
XII.) ;  produce  DE  to  F.  AG=AB,and  ADz=:  AC  (const.) ; 
hence  (Ax.  3)  DG=BC;  GIIcrAB.  The  rectangle  DH 
is  the  rectangle  of  GH  and  DG,  or  of  AB  and  BC.  AC 
=AD  or  BF,  and  CB=:BL:  hence,  by  adding  these 
equals,  AB=FL,  and,  being  sides  of  the  same  square,  KL 
=BC.  But  the  rectangle  KF  is  the  rectangle  contained 
by  FL  and  LK,  or  by  their  equals  AB  and  BC.  There- 
fore the  two  rectangles  DH  and  KF  are  the  rectangles 
contained  by  twice  AB  and  BC.  If  from  the  whole  fig- 
ure ACKLBHG  the  two  rectangles  DH  and  KF  be  sub- 
tracted, there  will  remain  the  square  AE.  But  the  whole 
figure  is  equal  to  the  two  squares  AH  and  CL.  Hence 
AC2=AB2+BC2-2AB.BC. 

Algebraically:  Let  ABr^a,  and  BC=5/  then  (a—hy 
—  a'^-\-h'^  —  2ah. 

Proposition  XVIII. — Theorem. 

The  rectangle  of  the  sum  and  difference  of  two  lines  is 
equal  to  the  difference  of  their  squares. 

Let  AB  and  AC  be  two  lines ;  then  will  the  rectangle 
of  their  sum  and  difference  be  equal  to  the  difference  of 
their  squares;  or  (AB-f-AC) .  (AB~AC)=AB2-AC2. 

On  AB  and  AC  construct  the  squares  AE  and  AH; 


BOOK   II. 


Gl 


prolong  BE  until  BL  is  equal 
to  AC,  draw  CK  equal  and  par- 
allel to  BL,  produce  CII  to  F, 
and  join  KL.  AB=rAD,  and 
AC=rAG;  hence  (Ax.  3)  CB  = 
DG.  BLz^AC  or  GH;  there- 
fore the  rectangle  GF  =z  the 
rectangle  CL.  EL=EB  +  BL, 
or  AB  +  AC  and  CB  or  KL=: 
AB-AC.  The  rectangle  EK 
is  the  rectangle  contained  by 
EL  and  LK.  Hence  the  rect- 
angle EK  is  also  the  rectangle 
contained  by  AB+ AC,  and  AB 
—AC.  Now  the  difference  of 
the  squares  of  AB  and  AC  is  the  rectangles  GF  and  CE. 
But  GF  was  before  proved  equal  to  CL:  hence  the  dif- 
ference of  the  squares  of  AB  and  AC  is  the  rectangle  EK. 
But  EK  is  the  rectangle  under  AB+AC,  and  AB— AC ; 
that  is,  (AB+AC).  (AB-AC)  =rAB2-AC2. 
Algebraically:  Let  AB=:a,  and  AC=b;  then  {a+b) 


a 

H 

A 

0 

Pkoposition  XIX. — Theorem. 

In  an  obtuse-angled  triangle  the  square  of  the  side  oppo- 
site the  obtuse  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  tico  sides  and  twice  the  rectangle  contained  by  the 
ba^e  and  the  distance  from  the  obtuse  angle  to  the  foot  of 
the  perpendicular  falling  on  the  base  produced. 

Let  ABC  be  an  obtuse-an- 
gled triangle;  then  will  the 
square  of  AC  be  equal  the  sum 
of  the  squares  of  AB  and  BC 
and  twice  the  rectangle  con- 
tained by  AB  and  BD. 

AD2z=:AB2+BD2  4-2AB  .  BD 
(Prop.  XYL),  to  each  add  DC^ ; 


62  BOOK  II. 

andAD24.DC2=AB2+BD2+DC2+2AB.BD.  ButAD^ 
4-DC2=AC2  (Prop.  XIIL),  and  BD2+DC2=BC2;  hence 
(by  substitution)  AC2=AB2+BC2+2AB.  BD. 

Proposition  XX. — Theoeem. 

In  an  acute-angled  triangle  the  square  of  a  side  oppo- 
site an  angle  is  equal  to  the  sum  of  the  squares  of  the 
other  tioo  sides  diminished  hy  twice  the  rectangle  con- 
tained by  the  base  and  the  distance  between  this  angle 
and  the  foot  of  a  perpendicular  let  fall  on  the  base. 

Let  ABC  be  an  acute-angled  ^ 

triangle ;  then  will  the  square 
of  AC  be  equal  to  the  sum  of 
the  squares  of  AB  and  BC 
diminished  by  twice  the  rect- 
angle contained  by  AB  and  ^^ 
BD. 

AD2=:AB2+BD2-2AB.BD  (Prop.  XVII.) ;  to  both 
add  DC2  and  AD2+DC2=AB2+BD2+DC2-2AB.BD. 
But  AD2+DC2=:AC2  (Prop.  XIIL),  and  BD2+DC2=: 
BC2.    Therefore  AC2= AB2+BC2-2AB .  BD. 

Proposition  XXI. — Theorem. 

If  one  side  of  a  triangle  be  bisected^  the  sum  of  the 
squares  of  the  other  two  sides  is  equal  to  twice  the  square 
of  half  the  side  bisected  and  twice  the  square  of  the  line 
drawyi  from,  the  point  of  bisection  to  the  opposite  angle. 

Let  AB  be  bisected  in  D 
by  the  line  DC ;  draw  EC 
perpendicular  to  AB.  Then 
will  AC2+BC2  be  equal  to 
2AD2+2DC2. 

For  AC2=AD2+DC2+ 
2AD.de     (Prop.   XIX.),  D 

and  BC2=:BD2+DC2-2BD .  DE  (Prop.  XX.).  Add  these 
two  equations  together,  and  AC2+BC2=AD24-DB2+ 
2DC2.     But  DB = AD ;  hence  AC2 + BC2 = 2 AD2 + 2DC2. 


BOOK   II. 


63 


Proposition  XXII. — Theorem. 

The  sum  of  the  squares  of  the  sides  of  a  parallelogram 
is  equal  to  the  sum  of  the  squares  of  tlie  diagonals. 

Let  ABCD  be  a 
parallelogram ;  then 
will  the  sum  of  the  x  x  jt 

squares  of  AB,  BD, 
CD,    and    AC    be 

equal   to    the    sum ,  

of  the    squares    of  -^ 

AD  and  BC,  or  AB2+BD2+CD2+AC2=AD2  4-BC2. 

The  diagonals  of  a  parallelogram  mutually  bisect  each 
other  (Prop.  IV.);  hence  AB^+AC2=:2EB2+2AE2,  and 
BD2+DC2=.2BE2+2DE2  (Prop.  XXL).  Add  these  equa- 
tions, and  AB2+AC24- BD2+DC2rzi4BE2 +2  AE2+2ED2; 
or  AB2+AC2+BD2+DC2=4BE2+4AE2.  But  since  the 
square  of  a  line  is  equal  to  four  times  the  square  of  its 
half,  AB2+AC2+BD2+DC2=AD2+BC2. 


The  foUoicing  are  Test  Examples  involving  the  First 
and  Second  JBooks  solved  or  demonstrated : 

1.   To  bisect  a  parallelogram  from  a  point  in  07ie  of  its 
sides. 

Let  ABCD  be  D  EC 

the  given  paral- 
lelogram, and  P 
the  given  point. 

On  the  line  DC 
lay  off  a  segment 
EC  equal  AP, 
and  join  the 
points  P  and  E.     The  line  PE  bisects  the  parallelogram. 

For,  draw  the  diagonal  AC*    The  two  triangles  ABC 


*  The  student  may  readily  invent  other  methods  of  demonstrating  this 
truth  by  drawing  different  hues. 


64  ELEMENTS  OF  PLANE  GEOMETRY. 

and  CDA  are  equal  (Prop.  III.).  In  the  two  triangles 
APO  and  CEO  the  angles  GAP  and  APO  are  equal  to 
OCE  and  CEO,  each  to  each  (Prop.  XXYH.,  Bk.  L),  and 
the  included  sides  AP  and  CE  are  equal  (const.).  Hence 
the  two  triangles  are  equal  in  all  their  parts  (Prop.  Y., 
Bk.  I.).  From  the  equal  triangles  ABC  and  CDA  sub- 
tract the  equal  triangles  APO  and  CEO,  and  there  will 
remain  the  quadrilateral  PBCO  equal  to  the  quadrilateral 
EDAO ;  and  to  these  equals  add  CEO  and  APO  respect- 
ively, and  the  sums  will  be  the  equal  quadrilaterals  PBCE 
andEDAP. 

2.  If  from  the  angular  points  of  the  squares  described 
upon  the  sides  of  a  right-angled  triangle  perpendiculars  he 
let  fall  upon  the  hypothenuse  produced^  they  will  cut  off 
equal  segments^  and  the  perpendiculars  will  he  together 
equal  to  the  hypothenuse. 

Let  ABC  be  the  given 
right-angled  triangle,  HC 
and  CG  the  given  squares, 
and  HE  and  GF  the  per- 
pendiculars upon  the  hy- 
pothenuse produced ;  then 
will  EA  be  eqtial  to  BF, 
and  HE  and  GF  will  be 
together  equal  to  AB. 
Draw  CD  perpendicular  to  AB.  In  the  two  triangles 
HEA  and  ADC,  the  angles  E  and  ADC  are  equal,  being 
right  angles.  The  sum  of  the  angles  CAD,  HAC,  and 
HAE  is  equal  to  two  right  angles,  and  the  sum  of  the 
three  angles  of  the  triangle  HEA  is  also  equal  to  two 
right  angles :  take  away  the  equal  angles  E  and  HAC, 
and  the  sum  of  the  angles  CAD  and  HAE  will  be  equal 
to  the  sum  of  the  angles  EHA  and  EAH ;  take  away  the 
common  angle  HAE,  and  the  angle  CAD  will  be  equal 
to  the  angle  EHA.  Hence  the  two  triangles  are  equian- 
gular. And  since  AH  is  equal  to  AC,  the  two  triangles 
are  equal  in  all  their  ^arts ;  EA  is  equal  to  CD,  and  HE 


BOOK   IT.  65 

is  equal  to  AD.  Similarly  it  can  be  proved  that  GF  is 
equal  to  DB,  and  BF  to  CD.  Since  EA  and  BF  are  each 
equal  to  CD,  they  must  be  equal  to  eacli  other ;  and  since 
HE  is  equal  to  AD,  and  G¥  equal  to  DB,  the  sum  of  HE 
and  GF  must  be  equal  to  the  sum  of  AD  and  DB  or  AB. 


TEST  EXAMPLES  IN  BOOK  II. 

1.  To  construct  a  quadrilateral  when  three  sides,  one 
angle,  and  the  sum  of  two  other  angles  are  given. 

2.  To  construct  a  quadrilateral  when  three  angles  and 
two  opposite  sides  are  given. 

3.  Prove  that  two  parallelograms  are  equal  when  they 
have  two  sides  and  the  included  angle  equal,  each  to 
each. 

4.  Prove  that  the  sum  of  the  diagonals  of  a  trapezium 
is  less  than  the  sum  of  any  four  lines  which  can  be  drawn 
to  the  four  angles  from  any  point  within  a  figure. 

5.  Prove  that  if  in  the  sides  of  a  square  four  points  be 
taken  at  equal  distances  from  the  angles,  the  lilies  joining 
these  points  will  form  anotl^er  square. 

6.  To  bisect  a  trapezfiiBi^  by  a  line  drawn  from  one  of 
its  angles. 

I.  Prove  that  if  lines  be^  drawn  from  the  extremities 
of  one  of  the  sides  of  a  trapezoid  to  the  middle  point  of 
the  opposite  side  (these  sides  not  to  be  the  parallel  sides), 
the  triangle  so  formed  will  be  half  the  trapezoid. 

8.  If  the  sides  of  an  equilateral  and  equiangular  penta- 
gon be  produced  to  meet,  the  angles  formed  by  these 
lines  are  together  equal  to  two  right  angles. 

9.  If  the  figure  be  a  hexagon,  prove  that  angles  formed 
as  above  will  be  equal  to  four  right  angles. 

10.  Prove  that  two  rhombi  are  equal  when  a  side  and 
an  angle  of  the  one  are  equal  to  a  side  and  an  angle  in 
the  other. 

II.  Prove  that  if  the  diagonals  of  a  quadrilateral  bisect 
each  other  at  right  angles,  the  figure  will  be  a  rhombus." 

12.  Prove  that  in  a  trapezoid  the  line  joining  the  mid- 
dle points  of  the  sides  which  are  not  parallel,  is  parallel 
to  the  parallel  sides.  ' 


ELEMENTS  OF  PLANE  GEOMETRY. 


1 3.  Prove  that  the  squares  of  the  diagonals  of  a  tra- 
pezium are  together  less  than  the  squares  of  the  four 
sides  by  four  times  the  square  of  the  line  joining  the 
points  of  bisection  of  the  diagonals. 

14.  Prove  that  if  squares  be  described  on  the  three 
sides  of  a  right-angled  triangle,  and  the  extremities  of 
the  adjacent  sides  be  joined,  the  triangles  so  formed  are 
equal  to  the  given  triangle  and  to  each-other. 

15.  Prove  that  if  lines  be  drawn  from  any  point  with- 
in a  rectangle  to  the  four  angles,  the  sums  of  the  squares 
of  those  lines  drawn  to  the  opposite  angles  are  equal. 

16.  Prove  that  the  sum  of  the  squares  of  the  diagonals 
of  a  trapezoid  is  equal  to  the  sum  of  the  squares  of  the 
two  sides  which  are  not  parallel,  and  twice  the  rectangle 
of  the  parallel  sides. 

17.  Prove  that  if  squares  be  described  upon  the  three 
sides  of  a  right-angled  triangle, 
and  the  extremities  of  the  square 
on  the  hypothenuse  be  joined  to 
the  extremities  of  the  adjacent 
lines  of  the  other  two  squares, 
the  sum  of  the  squares  of  the  two 
lines  thus  formed  will  be  equal 
to  five  times  the  square  of  the 
hypothenuse ;  or,  prove  that  GD^ 
+HK2=5AB2.  To  assist  the 
demonstration,  make  the  trian- 
gles GDL  and  HMK  right-angled. 


A  KEY  TO  THE  TEST  EXAMPLES  IN  BOOK  II. 

1.  See  Props.  XIV.,  XXX.  (Cor.l),  Bk.  L 

2.  See  same  Propositions. 

3.  See  Prop.  III.  and  Cor.  1. 

4.  SeeProp.Xn.,Bk.I. 

5.  By  constructing  the  figure  the  student  will  have  no 
difficulty  in  perceiving  the  equality  of  the  sides ;  and  by 
observing  the  relations  of  the  four  triangles  he  can  readily 
demonstrate  that  each  angle  of  the  small  square  is  a  right 
angle.  • 

6.  Bisect  CB  at  G,  and  through  G  draw  FE  parallel 


BOOK    II. 


67 


student  will  readily  prove  the 


to   AD.      From   the 
anc^le     ADC   .  draw 
DE :  tliis  line  DE  bi- 
sects  the   trapezfem 
ABCD.       See   Prop. 
III.    Prove  the  equal-   _ 
ity   of  the   triangles  A 
EGB  and  CGF.  .  The 
remainder. 

v.  Required  to  prove 
that  the  triangle  BEG 
is  half  the  trapezoid. 
Bisect  AD  at  E,  and 
draw  XY  parallel  to 
CB.     See  Prop.  IX.         A        X 

8.  See  Prop.  XXX.,  Bk.  I.,  Cor. 
1.  It  will  be  found  that  each  an- 
gle of  the  pentagon  is  z=  to  108° ; 
and  by  Prop.  XV.  each  exterior 
angle  is  12  .  The  student  can 
now  easily  finish  the  demonstra- 
tion. 

9.  Proved  in  a  similar  manner  \    / 
by  a  similar  construction. 

10.  This  theorem  is  easily  proved  * 

from  the  definition  of  a  rhombus,  and  by  drawing  di- 
agonals and  showing  the  equality  of  the  triangles  so 
formed. 

11.  Easily  demonstrated  by  means  of  the  propositions 
in  the  beginning  of  Book  I.  relating  to  the  equality  of 
triangles. 

12.  Through  F,  the  ^     ^ ___^    j^ 

middle  point  of  AC, 
draw  KE  perpendicu- 
lar to  AB:  produce 
CD  to  E,  and  draw 
CH  perpendicular  to 
FG.       It    is    easy    to 

prove  that  the  sum  of  

the   angles  HCD   and  -A-    ^  ^ 

CHG  is  equal  to  two  right  angles.  Hence  the  truth  of 
the  theorem. 


68 


KLEME^^TS    OF   PLANE    GEOMETRY. 


13.  Required  to  prove 
that  DB2+AC2.=AB2-f 
BC2+DC2+AD2_4FE2. 
See  Props.  XXI.  and 
XYI.,  Cor.  The  student 
commences  DC^+CB^rr: 
2DE2-f.2EC2;and,follow- 
^"  ing  up  this  line  of  demon- 
stration, the  truth  of  the  theorem  is  easily  established. 

14.  It  is  required  to  prove 
that  the  triangles  ABC,  AEF, 
GBL,  and  CMN  are  all  equal. 
A  glance  will  show  the  equality 
of  ABC  and  AEF.  BH  is  made 
N  equal  to  AB,  and  KC  to  AC. 
The  student  can  now  prove  that 
the  triangles  LHB  and  KCM 
are  each  equal  to  ABC  (Prop. 
IV.,  Bk.  I.) ;  and  by  Prop.  IX. 
he  can  demonstrate  that  LBH 
and  KCM  are  respectively  equal 
to  LBG  and  CMK 

15.  It  is  required  to  prove 
that  the  sum  of  the  squares  of 
AE  and  ED  is  equal  the  sum 
of  the  squares  of  CE  and  EB. 
DraV  the  diagonals  CB  and 
AD,  and  join  the  points  E  and 
F.     See  Props.  lY.  and  XXL* 

16.  It  is  required  to  prove 
that  AD2+BC2:^AC2+DB2 
+2(AB.CD). 

Erect  the  perpendiculars  AE 
and  BF,  and  produce  CD  to  E 
andF. 

*  The  students  who  can  solve  or  demonstrate  without  the  aid  of  the 
Key  should  be  encouraged  to  do  so ;  and  if  they  can  discover  other  and 
easier  methods,  so  much  the  better.  The  teacher  is  recommended  to 
make  his  reviews  in  Geometry  by  means  of  test  examples.  Indeed,  two 
or  three  of  these,  carefully  selected,  compel  a  general  review  of  the  pre- 
vious principles. 


D 


BOOK    II.  69 

AD2=AC2+CD24-2(CD.  CE),  Prop.  XIX. ; 
BC2:z.DB2+CD2+2(CD.DF);  and,  by  addition, 
AD2+BC2=AC2-fDB24-2CD24-2(CD.CE)4-2(CD.DF). 
Taking  FE  as  one  line  divided  into  three  parts,  and  Ct> 
as  another  line,  by  Prop.  XIV., 

EF.CD:=CD.EC+CD.CD+CD.DF; 
.•.AD24-BC2=AC2+DB2+2(EF.CD); 
AD24-BC2= AC2+DB2+2(AB.  CD). 
17.  The  figure  has  been  already  constructed.     Prove 
AN  =  to  AL,  and  NB=BM. 

DG2=AG2-f-AD2+2(AG.  AL), Prop.  XIX. 
The  student  who  has  attended  carefully  to  the  previous 
demonstration  will  find  no  difficulty  in  establishing  the 
truth  of  the  theorem. 


Q  '• 


.70  ELEMENTS    OF   PLANE    GEOMETRY. 


BOOK  III. 

DEFINITIONS. 

1.  An  Arc  is  any  portion  of  the  circumference  of  a 
circle. 

2.  A  Chord  is  a  line  joining  the  extremities  of  an  arc. 

3.  A  Tangent  is  a  line  without  a  circle,  which  touches 
it  in  one  point  only.j  ;        ,  / 

4.  A  Line  is  inscribed  in  a  circle  when  its  extremities 
terminate  in  the  circumference. 

5.  A  Secant  is  a  line  which  cuts  the  circumference  in 
two  points.  • 


6.  A  Segment  of  a  circle  is  that  part  of  it  bounded  by 
a  chord  and  the  arc  subtending  it. 

7.  A  Sector  of  a  circle  is  that  part  of  it  bounded  by 
two  radii  and  the  intercepted  arc. 

8.  An  Angle  is  inscribed  within  a  circle  when  the  vertex 
and  the  extremities  of  the  sides  are  in  the  circumference. 

9.  A  Polygon  is  inscribed  within  a  circle  when  all  its 
vertices  are  in  the  circumference. 

10.  The  Zone  of  a  circle  is  a  part  bounded  by  two  par- 
allel lines  and  the  intercepted  arcs. 

Proposition  I.-— Tiieore^i. 
Every  diameter  bisects  a  circle  and  its  circumference. 
Let  ACBD  be  a  circle ;  then  will  the  diameter  AB  bi- 
sect it  and  its  circumference. 


BOOK   III.  71 

Conceive  the  part  ADB  turned 
over  and  applied  to  the  part  ACB : 
it  must  exactly  coincide  with  it; 
for  if  it  does  not,  there  must  be 
points  in  either  portion  of  the  cir- 
cumference unequally  distant  from 
the  centre ;  which  is  contrary  to  the 
definition  of  a  circle.     Therefore  the  D 

two  parts  exactly  coincide  and  are  equal  (Ax.  10). 

Proposition^  II. — Theorem. 

A  line  perjyencUcidar  to  a  radius  at  its  extremity  is  tail- 
gent  to  the  circicmference. 

Let  the  line  CD  be  perpendicular  to 
the  radius  AB  at  its  extremity ;  then 
will  CD  touch  the  circumference  at 
one  point,  B,  only. 

From  any  point,  as  E,  in  the  line 
CD,  draw  the  line  AE  to  the  centre 
A.  Since  AB  is  perpendicular  to  CD, 
it  is  shorter  than  any  oblique  line,  AE 
(Prop.  XXXL,  Bk.  I).  Hence  the 
point  E  is  without  the  circle ;  and  in 
like  manner  it  can  be  proved  that  any 
other  point  in  the  line  CD  is  without  the  circle.  Hence 
CD  touches  the  circumference  in  one  point,  B,  only. 

Proposition  III. — Theorem. 

When  a  line  is  tangent  to  the  circumference  of  a  circle^ 
a  radius  drawn  to  the  point  of  contact  is  perpendicular  to 
the  tangent. 

Let  the  line  AB  be  tangent  to  the  circumference  of  a 
circle  at  the  point  D ;  then  will  the  radius  CD  be  perpen- 
dicular to  the  tangent  AB. 

For  the  line  AB  being  wholly  without  the  circumfer- 
ence, except  at  the  point  D,  it  follows  that  any  line,  as 


72 


ELEMENTS  OF  PLANE  GEOMETKY. 


(Prop.  XXXL,  Bk.  L). 


CF,  drawn  from  the  centre  C  to  meet 
the  line  AB  at  any  point  different 
from  D,  must  have  its  extremity,  F, 
without  the  circumference.  Hence 
the  radius  CD  is  the  shortest  line 
that  can  be  drawn  from  the  centre 
to  meet  the  tangent  AB ;  and  there- 
fore   CD    is   perpendicular    to   AB 


Pkoposition  IV. — Theoeem. 

If  a  line  drawn  through  the  centre  of  a  circle  bisect  a 
chord^  it  will  he  perpendicular  to  the  chord ;  or^  if  it  he 
perpendicular  to  the  chord^  it  icill  hisect  both  the  chord 
and  the  arc  of  the  chord. 

Let  the  line  CD,  drawn  from  the 
centre  of  the  circle,  bisect  the  chord 
AB  in  E ;  then  will  CD  be  perpendic- 
ular to  AB. 

Draw  the  radii  CA  and  CB.  In 
the  two  triangles  CAE  and  CBE  the 
sides  AC  and  CB  are  equal,  being  ra- 
dii of  the  same  circle  ;  AE  and  BE  are 
also  equal  (hyp.),  and  CE  is  common  to  both.  Hence  the 
two  triangles  have  three  sides  of  the  one  equal  to  three 
sides  of  the  other,  each  to  each,  and  are  equal  in  all  their 
parts  (Prop.  YIL,  Bk.  I).  Therefore  the  angle  AEC  is 
equal  to  the  angle  BEC  ;  and,  since  they  are  equal,  each 
must  be  a  right  angle,  and  the  line  CD  must  be  perpen- 
dicular to  AB  (def.  9). 

Again,  if  CD  be  perpendicular  to  AB,  then  will  the 
chord  AB  be  bisected  at  the  point  E,  and  the  arc  ADB 
be  bisected  at  the  point  D. 

For,  in  the  two  right-angled  triangles  AEC  and  BCE, 
the  hypothentise  AC  is  equal  to  the  hypothenuse  BC 
(def  19),  and  the  side  CE  is  common  to  both;  hence  the 
two  triangles  are  equal  in  all  their  parts  (Prop.  XXXII., 


BOOK    III.  73 

Bk.  I.) ;  the  side  AE  is  equal  to  the  side  BE,  and  the 
angle  ACE  is  equal  to  the  angle  BCE.  Tlien  apply  the 
sector  ACD  to  the  sector  DCB,  so  that  DC  will  be  com- 
mon ;  and  since  the  angle  ACD  is  equal  to  the  angle 
DCB,  the  line  AC  will  coincide  with  CB ;  and  since  the 
lines  AC  and  CB  are  equal,  the  point  A  must  fall  upon  B. 
The  sectors  must,  therefore,  coincide ;  for  if  they  did  not, 
some  point  of  the  arc  AD  Avould  fall  within  or  without 
the  circumference,  which  is  contrary  to  the  definition  of 
a  circle.     Hence  the  arc  AD  is  equal  to  the  arc  DB. 

PbOPOSITION  V. — PEOBLEJ^r. 

To  describe  a  circle  lohich  shall  pass  through  three 
given  points  not  in  the  same  straight  line. 

Let  A,  B,  and  C  be  three  points  not 
in  the  same  straight  line ;  it  is  required 
to  describe  a  circle  passing  through 
these  three  points. 

Join  the  points  A  and  B,  and  B  and 
C ;  bisect  AB  at  the  point  D,  and  BC 
at  the  point  E  (Prop.  IX.,  Bk.  L).  a:^ 
From  D  erect  the  perpendicular  DF,  and  from  E  erect 
the  perpendicular  EG  (Prop.  X.,  Bk.  I.) ;  the  point  of  in- 
tersection, H,  is  the  centre  of  the  circle  whose  circumfer- 
ence will  pass  through  A,  B,  and  C. 

The  perpendiculars  DF  and  EG  must  intersect ;  for  if 
they  do  not,  they  are  parallel,  which  is  impossible,  be- 
cause, if  DF  and  EG  were  parallel,  since  the  angles  HDB 
and  HEB  are  right  angles,  the  lines  AB  and  BC  muet  be 
the  secant  line,  and  form  one  and  the  same  straight  line, 
which  is  contrary  to  hypothesis.  In  the  two  triangles 
ADH  and  BDH,  AD  and  BD  are  equal  (const.).  DH  is 
common,  and  the  angles  ADH  and  BDH  are  equal,  being 
right  angles;  hence  the  two  triangles  have  two  sides  and 
the  included  angle  of  the  one  equal  to  two  sides  and  the 
included  angle  of  the  other,  each  to  each,  and  are  there- 

D 


74  ELEMENTS  OF  PLANE  GEOMETSY. 

fore  equal  in  all  their  parts  (Prop.  lY.,  Bk.  I.).  Hence 
AH  is  equal  to  BH;  and,  by  a  similar  demonstration, 
HC  can  be  proved  equal  to  BH;  and,  since  AH  and  CH 
are  each  equal  to  HB,  they  are  equal  to  each  other  (Ax. 
1).  Therefore  AH,  BH,  and  CH  are  radii  of  the  circle 
ABC,  which  passes  through  the  three  given  points  A,  B, 
and  C. 

Proposition  YI. — Problem. 
To  circwnscrihe  a  circle  about  a  given  triangle. 

Bisect  the  sides  AB  and  BC  in  the 
points  G  and  E,  and  from  these  points 
erect  perpendiculars  GO  and  EG. 
These  perpendiculars  must  intersect, 
because,  if  they  do  not,  they  must  be 
parallel,  and  if  parallel,  AB  and  BC, 
being  perpendicular  to  them,  would 
also  be  parallel,  which  is  impossible. 
Hence  GO  and  OE  must  meet.  Join  AO,  BO,  and  CO, 
and  from  O  let  fall  the  perpendicular  OF.  In  the  two 
right-angled  triangles,  AGO  and  BGO,  AG  and  BG  are 
equal  (const.),  and  GO  is  common.  "Fonpo  th<'  triangles 
are  equal  in  all  their  parts  (P^' -j  J--V"  ■^'^'  ^-i?  ^^  ^^ 
equal  to  OB.  By  comparing  the  two  triangles  BOE  and 
COE,  by  a  similar  demonstration  CO  can  be  proved  equal 
to  OB.  Hence  AO  and  OC,  each  being  equal  to  OB,  must 
be  equal  to  each  other.  Therefore  the  point  O  is  equi- 
distant from  the  points  A,  B,  and  C.  If,  then,  with  the 
point  O  as  centre,  and  OB  as  radius,  a  circumference  be 
described,  it  will  pass  through  the  points  A,  B,  and  C. 

Proposition  YII. — Theorem. 

Equal  chords  are  equally  distant  from  the  centre  /  and^ 
conversely^  chords  equally  distant  from  the  centre  are  equal. 

If  the  chord  AB  be  equal  to  the  chord  CD,  then  will 
they  be  equally  distant  from  the  centre. 

For  since  AB  is  equal  to  CD,  half  of  AB  must  be  equal 
to  half  of  CD.    Hence  AF  is  equal  to  CG.    We  al^o  have 


BOOK   III. 

HA  and  HC  equal,  and  the  angle  HFA 
equal  to  the  angle  HGC,  each  being  a 
right  angle.  Therefore  the  triangle 
HAF  is  equal  to  the  triangle  HGC 
(Prop.  XXXH.,  Bk.  I.),  and  consequent- 
ly HF  is  equal  to  HG. 

Conversely.     Let  AB  and  CD  be  any 
two  chords  equally  distant  from  the  centre  H:  then  will 
these  two  chords  be  equal  to  each  other. 

Draw  the  two  radii  HA  and  HC,  and  the  two  pei*pen- 
diculars  HF  and  HG,  which  are  the  equal  distances  of  the 
chords  from  the  centre  H.  The  two  right-angled  trian- 
gles HAF  and  HCG  have  the  sides  HA  and  HC  equal, 
being  radii,  the  side  HF  equal  to  HG,  and  the  angles  HFA 
and  HGC  right  angles.  Therefore  the  two  triangles  are 
every  way  equal  (Prop.  XXXH.,  Bk.  I.),  and  AF  is  equal 
to  CG.  But  AB  is  double  of  AF,  and  CD  is  double  of 
CG.     Hence  AB  is  equal  to  CD  (Ax.  6). 


Proposition  VHI. — Theorem. 

Equal  angles  at  the  centre  are  subtended  hy  equal 
chords ;  and^  conversely^  equal  chords  are  subtended  by 
equal  angles  at  the  centre. 

Let  the  angles  ACD^and  BCD  at  the 
centre  be  equal ;  then  will  the  chords 
AD  and  DB  be  equal. 

For  in  the  two  triangles  ACD  and 
BCD,  AC  and  CB  are  equal,  CD  is  com- 
mon, and  the  included  angles  ACD  and 
BCD  are  equal  (by  Hyp.).  Therefore 
AD  is  equal  to  BD  (Prop.  lY.,  Bk.  I). 

Conversely.  Let  the  chord  AD  be  equal  to  the  chord 
DB;  then  will  the  angle  ACD  be  equal  to  the  angle 
BCD. 

For  in  the  two  triangles  ACD  and  BCD  the  sides  AC 
and  CB  are  equal,  CD  is  common,  and  AD  is  equal  to 
DB  (by  Hyp.).  Therefore  the  angle  ACD  is  equal  to. 
the  angle  BCD  (Prop,  VH.,  Bk.  I). 


^Q  ELEMENTS  OF  PLAXE  GEOMETRY. 

Peoposition  IX. — Theoeem. 

I7i  the  same,  or  in  equal  circles,  equal  angles  at  the  centre 
intercept  equal  arcs, 

J±^  Let    ABK     and 

DEL  be  two  equal 
circles,  and  let  the 
angle  ACB  be  equal 
to  the  angle  DFE ; 
then  will  the  arc 
AB  be  equal  to  the 
TJ  JS  arc    DE.     Join   A 

and  B,  and  D  and  E. 

AC  is  equal  to  DF,  and  CB  is  equal  ^to  FE,  and  the  in- 
cluded angles  ACB  and  DFE  are  equal.  Therefore,  if 
the  sector  ACB  be  applied  to  the  sector  DFE,  the  point 
A  Avill  fall  upon  D,  and  B  upon  E,  and  the  arc  AB  must 
coincide  with  the  arc  DE,  for,  if  it  do  not,  some  point  of 
AB  must  fall  within  or  without  the  arc  DE,  which  is  con- 
trary to  the  definition  of  a  circle.  Hence  the  arcs  AB 
and  DE  are  equal. 

Cor,  1.  Equal  chords  are  subtended  by  equal  arcs,  and, 
conversely,  equal  arcs  by  equal  chords. 

Cor,  2.  Equal  arcs  are  subtended  by  equal  angles  at 
the  centre. 

Cor,  3.  An  angle  at  the  centre  is  measured  by  the  sub- 
tending arc.  All  the  angles  at  the  centre  are  together 
equal  to  four  right  angles,  or  to  360° ;  and  360^  is  the 
measure  of  the  circumference.  A  quadrant,  or  90°,  is  the 
measure  of  a  right  angle;  a  sextant,  or  60°,  is  the  meas- 
ure of  an  angle  of  an  equilateral  triangle. 

Peoposition  X. — Theoeem. 

An  angle  at  the  centre  is  double  an  ayigle  atthe  circum- 
ference  when  both  angles  stand  on  the  same  arc. 

Let  the  angle  ABD  at  the  circumference  stand  on  the 
arc  AD,  and  the  angle  ACD  at  the  centre  stand  on  tho 


BOOK   III. 


11 


same  arc  AD;    then  will  ACD   be 
double  ABD. 

For,  draw  the  diameter  BE.  AC 
and  BC  being  equal,  the  angle  A  is 
equal  to  the  angle  ABC ;  and  BC  and 
CD  being  equal,  the  angle  D  is  equal 
to  the  angle  CBD.  But  the  angle 
ACE  is  equal  to  the  sum  of  the  an- 
gles A  and  ABC,  and  the  angle  ECD  is  equal  to  the 
sum  of  the  angles  D  and  DBC  (Prop.  XXX.,  Bk.  L). 
Since  A  and  ABC  are  equal,  ACE  is  double  of  ABC ; 
and,  in  like  manner,  ECD  is  double  of  CBD.  Therefore 
the^sum  of  ACE  and  ECD  is  double  the  sura  of  ABC 
and  CBD;  or,  ACD  is  double  of  ABD. 

Co7\  1,  Since  ACD  is  measured  by 
the  arc  AD,  ABD  is  measured  by  half 
the  arc  AD ;  that  is,  an  inscribed  an- 
gle is  measured  by  half  the  subtend- 
ing arc. 

Scholium,  The  angle  might  have 
both  its  sides  on  either  side  of  the 
diameter.  In  this  case  it  may  be 
readily  shown  that  ACD  is  double 
of  ABD,  and  ECD  is  double  of  EBC. 
tion,  ACE  is  double  of  ABE. 

Cor,  2.  All  angles  inscribed  in  a  semicircle  are  right 
angles,  because  each  is  measured  by  half  the  semi-cir- 
cumference ;  that  is,  90°. 

Cor,  3.  All  angles  inscribed  in  an  arc  less  than  the 
semi-circumference  are  obtuse  angles,  because  each  is 
measured  by  half  of  an  arc  greater  than  1 80°. 

Cor,  4.  All  angles  inscribed  in  an  arc  greater  than  the 
semi- circumference  are  acute  angles,  because  each  is 
measured  by  half  of  an  arc  less  than  180°. 

Cor.  5.  All  angles  inscribed  in  the  same  arc  are  equal, 
because  each  has  the  same  measure. 

Cor,  6.  The  opposite  angles  of  an  inscribed  quadrilat- 


Then,  by  subtrac- 


^8  'elements  op  plane  geometry. 

eral  are  together  equal  to  two  right  angles,  because  half 
the  whole  circumference,  or  180°,  is  their  measure. 

.  Cor,  7.  The  sum  of  the  three  angles  of  an  inscribed 
triangle  is  equal  to  two  right  angles,  because  they 
are  measured  by  half  the  whole  circumference,  or  by 
180^ 

Proposition  XI. — Theorem. 

Parallel  ahords  intercept  equal  arcs, 

E  X  F      There  may  be  three  cases:   1st, 

^^— — \)oi]i  may  be  secants ;  2d,  both  may 

— -NP     be  tangents;   and,  3d,  one  may  be 

Ab  secant,  and  the  other  tangent. 

I         Let  AB  and  CD  be  parallel  se- 
/     cants ;  then  will  the  intercepted  arcs 
/      AC  and  DB  be  equal. 
^  Draw  the  radius  OK  perpendicu- 

^  ^  B:    lar  to  EF.     Then  the  arc  AKB  is 

bisected  at  K;  and  the  arc  CED  is  also  bisected  at 
K  (Prop.  ly.).  From  the  equals  AK  and  KB  subtract 
the  equals  CK  and  KD,  and  AC  will  remain  equal  to 
DB. 

Let  the  tangent  EF  be  parallel  to  the  tangent  GH; 
then  will  the  arc  LAK  be  equal  to  LBK.  The  arc  AK 
is  equal  to  the  arc  BK,  and  AL  is  equal  to  BL  (Prop. 
IV.) ;  and  by  adding  these  equals  LAK  is  equal  to 
LBK. 

Let  AB  be  a  secant,  and  EF  tangent.  The  radius  OK 
bisects  the  arc  AKB.     Therefore  AK  is  equal  to  KB. 

Proposition  XII. — ^Theorem. 

An  angle  formed  by  a  tangent  and  chord  is  measured 
by  half  the  arc  of  that  chord. 

Let  AB  be  a  tangent,  and  CD  a  chord  drawn  to  the 
point  of  contact ;  then  will  the  angle  BDC  be  measured 
by  half  the  intercepted  arc  DC.     Draw  the  diameter  ED. 


BOOK    III. 


'79 


The  angle  EDB  being  a  right  angle 
is  measured  by  half  the  semi- circum- 
ference— that  is,  by  the  arc  DCE  ;  and 
EDO  is  measured  by  half  the  arc  EC 
(Prop.  X.,  Cor.  1).  Now  if  the  angle 
EDC  be  taken  from  the  angle  EDB, 
and  the  arc  EC  from  the  arc  DCE, 
there  will  remain  the  angle  BDC 
measured  by  half  the  arc  CD;  and, 
by  addition,  the  angle  ADC  is  measured  by  half  DEC. 

Pkoposition^  Xin. — Theorem. 

The  angle  formed  hy  the  intersection  of  two  chords  is 
measured  by  half  the  sum  of  the  tico  intercepted  arcs. 

Let  the  two  chords  AB  and  CD  in- 
tersect each  other ;  then  will  the  angle 
AHD  be  measured  by  half  the  sum  of 
the  intercepted  arcs  AD  and  CB. 

Join  the  points  A  and  C.  The  an- 
gle AHD  is  equal  to  the  sum  of  the 
angles  HAC  and  HCA  (Prop.  XXX., 
Bk.  I.).  The  angle  HAC  is  measured 
by  half  the  arc  CB,  and  the  angle  ACH  is  measured  by 
half  the  arc  AD  (Prop.  X.  Cor.  1).  Therefore  the  angle 
AHD,  which  is  equal  to  the  sum  of  HAC  and  HCA,  is 
measured  by  half  the  sum  of  BC  and  AD. 

The  angle  CHB  is  vertical  to  AHD ;  therefore  CHB 
is  also  measured  by  half  the  sum  of  the  arcs  CB  and  AD. 

By  joining  AD  it  may  be  proved  in  a  similar  manner 
that  AHC,  or  its  equal  BHD,  is  measured  by  half  the 
sum  of  the  arcs  AC  and  BD. 


Peoposition  XIY. — Theorem. 

The  angle  formea^hy  two  secants  is  measured  hy  half 
the  difference  of  the  intercepted  arcs. 

Let  the  angle  B  be  formed  by  the  two  secants  AB 


80 


ELEMENTS  OF  PLANE  GEOMETRY. 


and  BC;  then  will  the  angle  B  be 
measured  by  half  the  difference  of  the 
arcs  AC  and  EF. 

Draw  ED  parallel  to  BC  (Prop. 
XXIX.,  Bk.  I).  The  angle  AED  is 
equal  to  the  angle  B  (Prop.  XXYIL, 
Bk.  I.).  But  the  angle  AED  is  meas- 
ured by  half  the  arc  AD.  Therefore 
the  angle  B  is  measured  by  half  the 
arc  AD.  AD  is  equal  to  the  differ- 
ence of  AC  and  DC.  But  DC  is  equal 
to  EF  (Prop.  XL).  Hence  AD  is  the  difference  of  the 
arcs  AC  and  EF.  Therefore  the  angle  B  is  measured  by 
half  the  difference  of  the  arcs  AC  and  EF. 


Proposition  XV, — Theorem. 


V" 


The  aiigle  formed  by  a  tangent  and  a  chord  drawn 
from  the  point  of  co7itact  is  equal  to  the  angle  inscribed 
i7i  the  alternate  segment  of  the  circle. 

Let  AB  be  a  tangent  and  EC 
a  chord,  forming  the  angle  CEB ; 
then  will  the  angle  CEB  be  equal 
to  any  angle,  as  D,  in  the  alter- 
nate segment  EDC. 

The  angle  CEB  is  measured  by 
half  the  arc  EC  (Prop.  XIL),  and 
-^  the  angle  D  is  measured  by  half 
the  same  arc  (Prop.  X.,  Cor.  1).  Therefore  the  angle  CEB 
is  equal  to  the  angle  D  (Ax.  1). 


Proposition  XVL — Problem. 

To  cut  off  a  segment  from  a  given  circle  which  shall 
contain  an  angle  equal  to  a  given  angle. 

Let  A  be  the  given  angle,  and  BECD  the  given  circle; 
then  it  is  required  to  cut  off  a  segment  Avhich  shall  con- 
tain an  angle  equal  to  A. 


BOOK   III. 


81 


Draw  the  tangent  FG. 
At  the  point  of  contact,  B, 
make  the  angle  EBG  equal 
to  A  (Prop.  XIV.,  Bk.  L). 
The  line  BE  cuts  off  the  re- 
quired segment,  BDE. 

For  the  angle  D  is  equal 
to  the  angle  EBG,  each  be- 
ing measured  by  half  the  , 
arc  BE.  But  the  angle  A  is  also  equal  to  EBG  (const.). 
Therefore  the  angle  D  is  equal  to  A ;  and  BDE  is  the 
required  segment. 

If  A  were  a  right  angle,  a  semicircle  would  be  the  re- 
quired segment. 

Pkoposition  XVII. — Peoblem. 

0)1  a  given  line  to  C07istruct  a  segment  of  a  circle  that 
shall  contain  an  angle  equal  to  a  given  angle. 

Let  A  be  the  given  angle,  and  BD 
the  given  line. 

At  the  point  B,  and  with  the  line 
BD,  make  an  angle,  CBD,  equal  to  A 
(Prop.  XIV.,  Bk.  I).  Bisect  BD  at 
E  (Prop.  IX.,  Bk.  1),  and  at  E  erect 
the  perpendicular  EF,  and  at  B  erect 
the  perpendicular  BG  (Prop.  X.,  Bk. 
I.).  F  is  the  centre  of  the  circle,  and 
BUD  the  required  segment. 

For,  in  the  two  triangles  BEF  and  DEF,  BE  is  equal 
to  DE  (const.),  and  EF  is  common,  and  the  included  an- 
gles BEF  and  DEF  are  equal.  Hence  BF  is  equal  to  DF 
(Prop.  IV.,  Bk.  I.),  and  the  circle  described  with  F  as 
centre  and  BF  as  radius  will  pass  through  the  point  D. 

Because  BG  is  perpendicular  to  BC  (const.),  BC  is  tan- 
gent to  the  circle  at  the  point  B  (Prop.  II.) ;  the  angle 
CBD,  made  by  a  tangent  and  a  chord,  is  equal  to  any 
angle  in  the  alternate  segment,  BHD,  of  the  circle  (Prop. 

D2 


82 


ELEMENTS  OF  PLANE  GEOMETRY. 


XY.).  But  A  is  equal  to  CBD.  Therefore  A  is  equal 
to  any  angle  inscribed  in  the  segment  BHD. 

Proposition  XVIII. — Problem. 

Within  a  given  circle  to  inscribe  a  triangle  equiangular 
to  a  given  triangle. 

Let  xyz  be  the  given  tri- 
angle, and  BED  the  given 
circle. 

Draw  AC  tangent  at  B :  at 
the  point  B  and  with  the  line 
BC  make  the  angle  CBD 
equal  to  the  angle  y,  and  at 
the  same  point  make  the  an- 
gle ABE  equal  to  x.  Join 
ED.  BED  is  the  required 
triangle. 
A  S  ei      For  the  angle  x  is  equal  to 

ABE  (const.),  and  the  angle  D  is  equal  to  ABE  (Prop. 
XY.).  Therefore  the  angle  x  is  equal  to  the  angle  D. 
It  can  be  proved  similarly  that  y  is  equal  to  E.  Hence 
the  remaining  angles  z  and  EBD  are  equal,  and  the  two 
triangles  are  equiangular. 


The  folloioing  Test  Examples  involve  the  Firsts  Second^ 
and  Third  Books, 

1.  Describe  three  circles  of  equal  diameters  which  shall 
touch  each  other. 

Take  any  line,  AB,  and  bisect  it  at  the  point  G  ;*  then, 
with  A  as  centre  and  AG  as  radius,  describe  the   cir- 


*  The  student  should  accurately  perform  every  construction  with  ruler 
and  compasses.  Unless  the  teacher  insists  upon  the  constant  use  of  the 
instruments  and  upon  accurate  measurements,  tho  study  of  Geometry 
will  be,  to  a  great  extent,  in  vain. 


BOOK    III. 


83 


cle  GKI ;  and  with  B  as 
centre  and  BG  as  radius, 
describe  the  circle  MHG. 
On  AB  describe  the  equi- 
lateral triangle  ABC 
then,  with  C  as  centre, 
and  half  of  AC  as  radi- 
us, describe  the  circle 
KHL.  It  is  evident 
that  the  three  circles, 
having  equal  radii,  have 
also  equal  diameters,  and  that  they  touch  each  other  in 
the  points  G,  K,  and  H. 

2.  In  an  equilateral  triangle  to  mscrihe  three  equal 
circles  lohich  shall  touch  each  other  and  the  three  sides 
of  the  triangle. 

Bisect  the  angles  A  and  B,* 
and  produce  the  bisecting  lines 
from  A  and  B  until  they  meet 
in  H.  Join  the  points  H  and 
C.  It  is  evident  that  AHB 
is  an  isosceles  triangle  (Prop. 
Y.,  Bk.  I.) ;  and  the  two  trian- 
gles AHC  and  BHC  have  two 
sides  and  the  included  angles 
equal.     Therefore  (Prop.  IV.,  A  '  ^B 

Bk.  I.)  the  angle  ACH  is  equal  to  BCH,  and  the  trian- 
gles are  isosceles.  Again,  bisect  angles  at  A,  B,  and 
C,  of  the  three  isosceles  triangles,  by  the  lines  AD,  DC, 
CE,  EB,  BF,  and  FA.  The  points  of  intersection,  D,  E, 
and  F,  will  be  the  centres  of  the  three  equal  circles. 
From  the  previous  propositions  of  the  First  and  Third 
Books  the  student  will  have  no  difficulty  in  proving  the 

*  The  student  will  perceive  that  every  operation  has  been  accurately 
performed.  This  doing  the  work  frequently  makes  that  clear  which  was 
before  obscure. 


84  ELEMENTS  OF  PLANE  GEOMETRY. 

equality  of  the  inscribed  circles.     He  has  only  to  prove 
that  their  radii  are  equal. 

3.  If  from  each  extremity  of  any  nicniber  of  equal  arcs 
lines  he  draion  through  two  given  points  in  the  opposite 
part  of  the  circumference  and  produced  till  they  meet^  the 
angles  formed  by  these  lines  will  be  equal 

^^.^-.^^^^  LetABandBCbe 

A/^  ^\  equal  arcs,  and  F  and 

2ti  Z^"'^^-....,^^^^        \  E  two  points  in  the  op- 

-g/     ^^  '^"--^"^^^^^"^^ -^  posite  part  of  the  cir-- 

m/_^^  cumference,  through 

^Ssc-"^ — '    '      y^^        ^^-^^^==::5s»^X    which   let    the    lines 

^>^^^_^y^  AFI,  BEI,  BFK,  and 

CEK  be  drawn ;  the 
angles  at  I  and  K  will  be  equal. 

Through  the  point  E  draw  Em  parallel  to  AF,  and  E?z 
parallel  to  BF.  Since  Em  is  parallel  to  AFI,  the  angle 
BEm  is  equal  to  the  angle  I  (Prop.  XXYIL,  Bk.  I.),  and, 
for  a  similar  reason,  the  angle  CE^i  is  equal  to  the  angle 
K.  Parallel  chords  intercept  equal  arcs  (Prop.  XL). 
The  arcs  Am  and  B?2  are  both  parallel  to  FE;  therefore 
they  are  equal  to  each  other.  But  AB  is  equal  to  BC 
(hyp.),  and  if  the  former  equals  be  taken  from  the  latter, 
the  arc  Bm  will  be  equal  to  the  arc  Qn,  Therefore  the 
angle  BEm  is  equal  to  the  angle  CE^^  (Prop.  IX.,  Cori$). 
But  I  and  K  were  before  proved  equal  to  these  two  an- 
gles, each  to  each]  therefore  they  are  equal  to  each  other. 


TEST  EXAMPLES  IN  BOOK  III. 

1.  Through  two  given  points  to  draw  a  circumference 
of  given  radius.  (The  radius  must  be  greater  than  half 
the  distance  between  the  two  points.) 

2.  Draw  a  tangent  to  a  given  circle  parallel  to  a  given 
hne. 

3.  Describe  a  circle  of  given  radius  tangent  to  a  given 
line  at  a  given  point. 


BOOK   III.  85 

4.  Describe  a  circle  of  given  radius  touching  the  two 
sides  of  a  given  angle. 

5.  Prove  that  if  a  circle  be  described  on  the  radius  of 
another  circle,  any  straight  line  drawn  from  the  point 
where  they  meet  to  the  outer  circumference  is  bisected 
by  the  interior  one. 

6.  Prove  that  if  two  circles  cut  each  other,  and  from 
either  point  of  intersection  diameters  be  drawn,  the  ex- 
tremities of  these  diameters  and  the  other  point  of  inter- 
section shall  be  in  the  same  straight  line. 

7.  Describe  a  circumference  which  shall  be  embraced 
between  two  parallels  and  pass  through  a  given  point 
within  the  parallels. 

^^  8.  Find  in  one  side  of  a  triangle  the  centre  of  a  circle 
which  shall  touch  the  other  two  sides. 

9.  -Through  a  given  point  on  a  circumference,  and  an- 
other given  point  without,  to  describe  a  circle  touching 
the  given  circumference. 

10.  In  the  diameter  of  a  circle  produced,  to  determine 
a  j'oint  from  which  a  tangent  drawn  to  the  circumference 
£  >rJl  be  equal  to  the  diameter. 

'  1.  Prove  that  in  a  quadrilateral  circumscribing  a  cir- 
(  '^  the  opposite  sides  are  equal  to  half  the  perimeter. 
12.  Prove  that  if  the  opposite  angles  of  a  quadrilateral 
equal  to  two  right  angles  a  circle  may  be  described 
•     nit  it. 

■  3.  Describe  a  circle  of  given  radius  touching  two 
.  ren  circles. 

i  4.  In  a  given  circle  to  inscribe  a  right  angle,  one  side 
'  which  is  given. 

15.  In  a  given  circle  to  construct  an  inscribed  triangle 
.  given  altitude  and  vertical  angle. 

16.  Prove  that  if  an  equilateral  triangle  be  inscribed 
ill  a  given  circle,  the  square  described  on  a  side  is  equal 

v»  three  times  the  square  described  on  the  radius. 

i  7.  Inscribe  a  square  in  a  given  right-angled  isosceles 
1  i  angle. 

18.  Inscribe  a  square  in  a  given  quadrant  of  a  circle. 
,1 0.  Find  the  centre  of  a  circle  in  which  two  given  lines 
ine^^ting  in  a  point  shall  be  a  tangent  and  a  chord. 

'20.  Inscribe  a  square  in  a  circle,  and  circumscribe  a 
circle  with  a  square. 


86 


ELEMENTS  OP  PLANE  GEOMETRY. 


21.  Inscribe  in  a  circle  a  regular  hexagon;  also  an 
equilateral  triangle. 

22.  Describe  a  circle  the  circumference  of  which  shall 
pass  through  a  given  point  and  touch  a  given  straight 
line  in  a  given  point. 

23.  If  a  circle  be  inscribed  in  a  right-angled  triangle, 
the  difference  between  the  sum  of  the  two  sides  contain- 
ing the  right  angle  and  the  hypothenuse  is  equal  to  the 
diameter  of  the  circle. 


A  KEY  TO  THE  TEST  EXAMPLES  IN  BOOK  III. 

1.  See  Props.  IX.  and  X.,  Bk.  I. 

2.  See  Props.  XI.,  XXIX.,  Bk.  I.  and  III.,  Bk.  III. 

3.  See  Props.  X.,  Bk.  I.,  and  III.,  Bk.  III. 

4.  See  Prop.  VIII.,  Bk.  I. :  every  point  of  the  bisectrix 
of  an  angle  is  equally  distant  from  the  sides.  If  perpen- 
diculars be  let  fall  from  any  point,  two  right-angled  tri- 
angles will  be  formed  having  two  angles  and  the  included 
side  of  the  one  equal  to  two  angles  and  the  included  side 
of  the  other,  each  to  each. 

5.  The  circle  ABD  is  described 
upon  the  radius  AB  of  the  circle 
AEC :  it  is  required  to  prove  that 
the  line  AC  is  bisected  at  the  point 
D.  Produce  AB  to  E  and  draw 
DB.  The  angle  ADB  being  in- 
scribed in  a  semicircle  is  a  right 
angle  (Prop.  X.).  Hence  DB, 
drawn  from  the  centre,  B,  of  the 
large  circle,  is  perpendicular  to  the  chord  AC,  and  bisects 
it  (Prop.  IV.). 

6.  From  A  draw  the  diameters  AC  and  AD,  and  from 

B  draw  BC  and  BD.    Join 
AB.     See  Prop.  X. 

7.  Take  two  points  in 

the  parallel  lines,  which, 

with  the  given  point,  shall 

^  not  be  in  the  same  straight 

line.     See  Prop.  V. 


f/H.    A, 


BOOK    111. 


87 


8.  See  Test  Example  4. 

9.  Join  the  two  points. 
See  Props.  IX.  and  L,  Bk.  I. 

10.  From  A  draw  AD 
perpendicular  and  equal  to 
AB.  From  the  centre  O 
draw  OD ;  and  from  C 
draw  CE  perpendicular  to 
OD,  meeting  BA  produced 
at  E.  By  comparing  the 
two  triangles  the  student 
can  readily  prove  that  CE 
z:=AD=BA. 

11.  Draw  the  lines  OB,  00, 
OE,  and  OG  perpendicular  to 
the  four  sides  of  the  quadrilat- 
eral ;  and  join  OA,  OD,  OF, 
and  OH.  By  comparing  the 
triangles  there  will  be  no  diffi- 
culty in  proving  AC-f  CD  = 
AB+DE,  and  HG+GF=FE 
+HB. 

12.  See  Prop.  X.,  Cor.  1. 
The  sum  of  the  tw^o  angles  will  require  (when  they  are 
inscribed)  the  whole  circumference  to  measure  them. 
See  also  Prop.  V. 

13.  Draw  two  circles,  respectively  equal  to  the  given 
circles,  in  such  a  manner  that  the  distance  between  them 
shall  be  double  the  given  radius.  Bisect  this  distance, 
and  with  the  point  of  bisection  as  centre,  and  half  the 
distance  as  radius,  describe  the  required  circle. 

14.  See  Prop.  X.,  Cor.  2. 

15.  Draw  an  inscribed  angle  equal  to  the  given  angle. 
It  is  evident  that  the  altitude  is  al- 
ways limited  by  the  measure  of  the 
inscribed  angle. 

16.  The  angle  ADE  is  double 
ABE,  or  is  =  to  ABC  or  ACB.  But 
the  angles  C  and  E  are  equal  (Prop. 
X.  and  Cor.  1).  ,Hence  the  angle 
ADE  =  the  angle  E.  Therefore 
AE=AD.      The    square   of  EB   is 


/^;  ^i     /_ 


88 


ELEMENTS  OF  PLANE  GEOMETRY. 


equal  to  AE^+AB^,  or  4  times  ED^^AE^+ABs.  SuV 
tract  AE^  from  one  side,  and  its  equal  DE^  from  the 
other,  and  3  times  DE2=AB2. 

17.  Trisect  the  hypothenuse  (see  Test  Ex.,*Bk.  I.),  and 
from  the  points  of  trisection  erect  perpendiculars  to  the 
other  two  sides.  Connect  the  points  in  the  two  sides  cut 
by  these  perpendiculars.  The  middle  portion  of  the  hy- 
pothenuse, the  perpendiculars,  and  the  connecting  line 
form  the  square  required. 

18.  Bisect  the  right  angle  to  the  quadrant.  The  re- 
mainder is  quite  simple. 

19.  See  Props.  IV.  and  XII.,  Bk.  III. 

20.  The  student  should  solve  this  without  any  help. 

21.  The  chord  of  60  degrees  is  equal  to  the  radius  of 
the  circle. 

22.  Let    AB    be    the    given 
straight  line,  C  the  given  point 
in  which  the  circle  is  to  touch  it, 
and  D  the  point  through  which 
it  must  pass.    Draw  CO  perpen- 
dicular to  AB.      Join  CD;  and 
at  the  point  D  make  the  angle 
CDO=DCO:  the  intersection  of 
the  lines  CO  and  DO  is  the  cen- 
tre of  the  circle  required.     Since  the  angle 
DCO=:CDO,  CO=T>0.     Therefore  a  circle 
jdescribed  from  the  centre  O,  at  the  distance 
OD,  will  pass  through  C  and  touch  the  line 
AB  in  C,  because  OC  is  perpendicular  to  AB. 
23.  Find   the    centre   O.     Join    OD   and 
OE.     The  angles  B,  D,  and  E  are  right  an- 
gles ;  OD  and  OE  are  equal.     Hence  OB  is 
a  square.     In  Test  Example  11  it  is  shown 
\  /\    that  FC  =  CE,  and  that  AF=:AD.     Hence 

Vj^ZA  AC=EC+AD,orAB+BC~BE+BD.  But 
C  BE+BD=:the  diameter. 


BOOK   IV.  89 


BOOK  lY. 

DEFINITIONS. 

1.  A  LESS  magnitude  or  quantity  is  a  measure  ol^  a 
greater  magnitude  or  quantity  when  the  less  is  exactly 
contained  a  certain  number  of  times  in  the  greater. 

2.  A  greater  magnitude  is  a  multiple  of  a  less  when 
the  greater  is  measured  by  the  less;  that  is,  when  the 
greater  contains  the  less  an  exact  number  of  times. 

3.  Ratio  is  the  relation  of  two  magnitudes  of  the  same 
kind,  the  one  to  the  other ;  or  it  is  the  quotient  arising 
from  the  division  of  the  one  by  the  other. 

4.  PropoTtioii  is  an  equality  of  ratios. 

5.  Three  quantities  are  in  projDortion  when  the  first  is 
to  the  second  as  the  second  to  the  third.  Four  quantities 
are  in  proportion  when  the  first  is  to  the  second  as  the 
third  to  the  fourth. 

For  example,  A:B::B:C.  Let  A=2,B=4,and  C= 
8.     Then  2:4::4:8;  that  is, 

-  =- ;  the  equality  of  ratios. 

4     8 

In  this  example  B,  or  4,  is  a  mean  proportional  between 
A  and  C,  or  between  2  and  8. 

Again,  let     A=3,  B=4,  C=6,  and  D=:8. 
Then       A:B  ::C:D,  or,  3  :  4  ::6  :  8;  that  is, 

-=-;  an  equality  of  ratios. 

The  first  terra  is  called  the  antecedent,  the  second  the 
consequent,  the  third  the  antecedent,  and  the  fourth  the 
consequent,  and  so  on. 

6.  Magnitudes  are  in  continued  proportion  when  they 
have  a  common  ratio ;  as,  1  is  to  2  as  2  to  4,  as  4  to  8, 
and  so  on.     Here  the  common  ratio  is  \. 


90  ELEMENTS    OF    PLANE    GEOMETRY. 

7.  The  first  and  fourth  terms  are  called  the  extremes^ 
and  the  second  and  third  are  called  the  means,  ^ 

8.  An  Inverse  Proportion  is  where  the  antecedent  is 
made  the  consequent,  and  the  consequent  the  antece- 
dent.    Thus,  if  2  :  3  : :  6  :  9,  then,  by  inversion,  3  :  2  : :  9  :  6. 

9.  Alternate  Proportion  is  where  antecedent  is  com- 
pared with  antecedent,  and  consequent  with  consequent. 
Thus,  if  2 :  3  : :  6  :  9,  then,  by  alternation,  it  will  be  2:6:: 
3:9. 

10.  A  Compound  Proportion  is  where  the  sum  of  the 
antecedent  and  consequent  is  compared  either  with  the 
antecedent  or  consequent.  Thus,  if  2 : 3  : :  6 :  9,  then,  by 
composition,  2  +  3  :  2  ::  6  +  9  :  6. 

11.  ^  Divided  Proportion  is  where  the  difference  of  the 
antecedent  and  consequent  is  compared  with  either  the 
antecedent  or  consequent.  Thus,  if  2  :  3  : :  6 :  9,  3 — 2 :  3  : : 
9  —  6:9. 

Proposition  I. — Theorem. 

Equimultiples  of  any  two  magnitudes  have  the  same 
ratio  as  the  magnitudes  themselves. 

Let  A  and  B  be  ^ny  two  magnitudes,  and  mK  and  wB 
any  equimultiples  of  them  {m  being  any  quantity  what- 
ever), then  will  mK  and  mB  have  the  same  ratio  as  A 
and  B,  or  A :  B : :  mK :  mB. 

^  mK    A 

^"^"  ^=B- 

Let         A=2  B4  and  m  any  number,  say  6  : 

6x2  2  ^,  ^  .  '12  2 
- — r=T;  thatis,— -=-. 
6x4     4'  '24 

Proposition  IL — Theorem. 

If  four  m^agnitudes  are  proportional^  the  product  of  the 
extremes  is  equal  to  the  product  of  the  means. 

Let  the  four  magnitudes  A,  B,  C,  and  D  form  the  pro- 
portion A :  B : :  C :  I),  then  will  A  x DrrB  x  C 


BOOK   IV.  91 

Since  A:B::C:D,g=^(Def.4). 

Multiply  these  equals  by  BD,  and  we  have 

AD=BC. 
Let  A=2,  B=:3,  C  =  6,  and  D  =  9, 

2:3::6:9,and  2x9  =  3x6. 

Proposition  III. — Theorem. 

If  the  product  of  two  magnitudes  he  equal  to  the  prod- 
uct of  two  other  magnitudes^  they  wiU  constitute  a  pro- 
portion in  which  either  two  ^nagnitudes  will  be  the  ex- 
tremes^ and  the  other  two  the  means. 

Let  A  X  B=C  X  r>.     Dividing  both  members  by  B  x  D, 

and  we  have  ^^ — TT==n — ^ ;  striking  out  the  common  fac- 
BxD    BxD'  ^ 

A    C 

tors,  and  j^=^>  or  A :  D : :  C :  B. 

Let    A=2,B=:9,C  =  6,andD  =  3;  2x9=3x6, 
2x9'     ^X6     2     6        o    o     «    n 
^^=9-3^=3  =  9'^^ '^'^••^^'- 

Proposition  IV. — ^Theorem. 

If  four  magnitudes  are  proportiojial^  they  will  be  in  pro- 
portion by  alternation. 

Let  four  magnitudes,  A,  B,  C,  and  D,  be  in  proportion, 
as  A :  B : :  C :  D,  then  will  they  be  alternately  in  propor- 
tion as  A:C::B:D. 

A    C 

For  since  A :  B : :  C ;  D,  -^ = ^r.     Multiplying  both  mem- 

-        ,     B        ,        AxB    CxB      ^    ., . 
bers  by7r,we  have  -rz — 77=7^ — t^.     btrikmg  out  common 
•^  C  BxC    CxD  ^ 

A    B 

factors,  we  have  -7^=:pr,  or  A :  C : :  B :  D. 

Let  A=3,  B=:6,  C=4,  and  D=8, 

3:6::4:8,  or  3  :4::6:8. 


92  ELEMENTS  OF  PLANE  GEOMETRY. 


Proposition  V. — Theorem. 

If  four  mag^iitiides  are  in  proportion^  they  loill  he  in 
proportion  inversely. 

Let  four  magnitudes,  A,  B,  C,  and  D,  be  in  proportion, 
as  A :  B : :  C :  D,  then  will  they  be  in  proportion  inverse- 
ly, as  B :  A ::  D :  C. 

Since  A:  B ::  C  :  D,  ^=:^,  or  B  :  A  ::D :  C. 

Let  A=4,B=8,Cr=6,  andD=:12;  4:8::6:12,or  8:4 
::12:6. 

Proposition  VI. — Theorem. 

If  four  magnitudes  he  proportional^  and  four  other  mag- 
nitudes proportional^  having  the  antecedents  the  same  in 
hoth^  the  consequents  will  he  proportional. 

Let  A  :  B : :  C :  D  be  four  proportional  magnitudes,  and 

A:m\\C:x  four  other  proportional  magnitudes,  having 

the  antecedents  A  and  C  the  same  in  both,  then  will  the 

consequents  be  proportional,  B :  m : :  D :  a;. 

A     C 
A:B::C:Dgives^=:^, 

and  A:m\\(j:x  c?ives  — r=  — . 

°  7)1       X 

Divide  the  second  equation,  member  by  member,  by 

,     ^  ^        ^        AB     CD       B     D 

the  first,  and  we  have  -. — =77—5  or  — =— ,  or 
■  Am     Kjx        m     X 

B :  m : :  D  :  oj. 

Let  A=:2,  B=4,  C=6,  D  =  12,  m=3,  and  i^=9;   2:4 

::  6: 12,  and  2:3::  6: 9.     Hence  4:  3::  12:  9. 

Proposition  Vlt — Theorem. 

If  four  inagnitudes  he  proportional^  they  will  he  iri  pro- 
portion  hy  co'inposition  and  division. 

Let  A :  B  : :  C :  D  be  a  proportion  ;  then  will  A+B :  A 
::  C+D :  C,  or  A—B :  A ::  C— D :  C  be  also  proportional. 


BOOK    IV.  93 

For  BxC=:AxD;  by  adding  both  members  of  this 
equation  to,  and  subtracting  them  from,  A  x  C,  we  have 
AxC±BxC=:AxC±:AxD; 
(A±B)xC  =  (C±D)xA; 
A±B:A::C±D:C. 
Let  A=2,  B  =  4,  0  =  5,  and  D=10 ; 

2  +  4:2::5  +  10:5; 
2-4:2::5-10:5; 
6:2::15:5; 
—2:2::— 5:5. 

Peopositiox  VIII. — Theorem. 

Of  four  proportional  rnagnitudes^  if  any  equimultiples 
of  the  antecedents  lohatever^  and  any  equimultiples  of  the 
co7isequents  he  taJcen^  the  resulting  magnitudes  will  be  pro- 
portional. 

Let  A,  B,  C,  and  D  be  in  proportion  ;  as  A :  B  : :  C :  D, 
and  r  be  an  equimultiple  of  the  antecedents,  and  s  an 
equimultiple  of  the  consequents ;  then  will  rA,  rB,  sC, 
and  sT)  be  in  proportion,  as  rA :  rB : :  sC :  sD. 

For,  since  A:B::C:D,AxD=:BxC.  Multiply  both 
members  by  r  x  5,  and  we  have 

T'AxsD^rBxsC; 
rA  :  rB::5C  :  5D. 
Let  A  =  3,B=::6,C  =  8,andD  =  16,r=:4,and5=3; 
3:6::  8: 16; 

3x4:6x3::8x4:16x3; 
12: 18::  32:  48. 

Proposition  IX. — Theorem. 

If  there  he  four  proportional  magnitudes^  and  the  two 
consequents  he  either  augmented  or  diminished  hy  magni- 
tudes that  have  the  same  ratio  as  the  respective  antecedents^ 
the  residts  and  the  antecedents  will  still  be  proportional. 

Let  A:B::C:D,  and 

A:  C  ::m:ny 
then  will  A\Q\\B±m\T>±n. 


04  ELEMENTS    OF    PLANE    GEOMETRY. 

For,  since        A :  B : :  C :  D,  A  x  D = B  x  C ; 
and,since  A:C::m:;^,  Axnz=:Cxm. 

By  addition  and  subtraction  of  equals, 

AxD±Ax^=BxCdbCxm. 

By  factoring,  A  x  (D  db  n)  =  C  x  (B  zt  m) . 

Therefore        A :  C : :  B  ±  m :  D  zb  72 . 

Let  A=2,  Bzz:4,  C=:3,  and  Dr=:6.  Then  increase  the 
two  consequents  of  the  proj)ortion  2  :  4  : :  3  :  6  by  1  and 
1^,  and  we  have  2 :  3  : :  5  :  7^ ;  or  diminish,  and  w^e  have 
2:3::3;4|. 

Proposition  X.— Theorem. 

If  any  ^nagnitiides  he  proportional^  any  like  powers  or 
roots  of  them  will  he  proportional. 

Let  A:B::C:D; 

then  will  A^rB^-C^iD^, 

or  A3:B3::C3:D3, 

or  VA:  VB::  v/C:  VS. 

AxD=:BxC;  squareand  A^xD^^rB^xG^;  cube  and 

A^  X  r>^=B^  X  C^ ;  extract  the  square  root,  and  -/A  x  D=: 

VBxC;  that  is,  A2 : B2 : : C2 : D2 ; 

A3 :  B3 : :  C^ :  D^ ;  or  VA :  V^^: :  yC :  VD. 
Let       A=:2,B=4,C  =  3,andD=:6, 
and  2:4:: 3: 6, or  22^42:: 32; 62; 

4:16::9:36,or23:43::33:63; 

8 : 6  4 : :  2  7 : 2 1 6 ,  o r  V  2 :  V'l : :  -v/i  :  V6 . 

It  will  be  found  that  by  substituting  numerical  values 
for  the  magnitudes  the  propositions  of  the  Fourth  Book 
can  be  demonstrated  with  great  facility ;  and  young  stu- 
dents can  be  taught  to  comprehend  them  without  diffi- 
culty. Beginners  do  not  readily  perceive  that  A  and 
B,  m  and  n,  and  x  and  y  have  a  geometrical  meaning. 
However,  when  A  is  2  and  B  4,  they  see  instantly  the 
relation  of  2  to  4,  although  the  relation  of  A  to  B  may 
be  very  vague. 


BOOK    V. 


9S 


BOOK  V. 

DEFINITIONS. 

1.  Similar  polygons  are  those  which  have  their  angles 
equal,  each  to  each,  and  the  sides  about  the  equal  angles 
proportional. 

2.  In  similar  polygons  the  sides  adjacent  to  the  equal 
angles  are  called  homologous  sides,  and  the  angles  them- 
selves are  called  homologous  angles. 

3.  Two  sides  of  one  polygon  are  said  to  be  reciprocally 
proportional  to  two  sides  of  another  when  one  of  the  sides 
of  the  first  is  to  one  of  the  sides  of  the  second  as  the  re- 
maining side  of  the  second  is  to  the  remaining  side  of  the 
first. 

4.  The  altitude  of  a  triangle  is  the  perpendicular  from 
the  vertex  to  the  base,  or  the  base  produced.  The  alti- 
tude of  a  parallelogram  is  the  perpendicular  between 
two  opposite  sides.  The  altitude  of  a  trapezoid  is  the 
perpendicular  between  the  two  parallel  sides. 

5.  Area  denotes  the  superficial  contents  of  a  figure. 

A  and  a,  B  and  h^  C  and  c,  are  homologous,  each  to 


b   A.  B 

each.     AB  is  homologous  to  ah^  BC  to  5c,  and  AC  to  ac. 
The  triangles  are  similar,  and  the  sides  are  proportional. 


96 


ELEMENTS    0^   PLANE    GEOMETRY. 


D           i 

A             1 

C 

B 

Proposition  I. — Theorem. 

Two  rectangles  of  the  same  altitude  are  to  each  other  as 
their  bases. 

Let  ABCD 
and  AFGD  be 
two  rectangles 
having  the 
common  alti- 
tude AD ;  then  will  they  be  to  each  other  as  their  bases 
ABandAF. 

First,  Suppose  the  bases  AB  and  AF  to  be  commensu- 
rable ;  as,  for  example,  suppose  they  are  to  each  other  as 
5  to  3.  If  AB  be  divided  into  5  equal  parts,  AF  will  con- 
tain 3  of  these  parts.  At  each  point  of  division  draw 
lines  perpendicular  to  the  base,  forming  5  rectangles, 
which  will  be  equal,  since  they  have  equal  bases  and 
altitudes  (Prop.  VII.,  Bk.  II.). 

Since  the  rectangle  ABCD  contains  5  of  these  rectan- 
gles, and  the  rectangle  AFGD  contains  3,  it  follows  that 

ABCD:  AFGD  ::5:  3; 
but  AB:AF::5:3; 

therefore  ABCD  ;  AFGD : :  AB :  AF. 

Second,  Suppose  AB 
and  AF  to  be  incommen- 
surable ;     still    ABCD : 


AFGD::AB:AF. 

For  if  this  proposition 
be  not  true,  the  first  three 
terms  remaining  the  same, 
the  fourth  term  will  be  greater  or  less  than  AF.  Suppose 
it  to  be  greater,  then  ABCD :  AFGD : :  AB  :  AL.  Divide 
the  line  AB  into  equal  parts,  each  of  which  shall  be  less 
than  FL.  There  will  be  at  least  one  point  of  division, 
as  at  H,  between  F  and  L.  Through  this  point  H  draw 
the  perpendicular  UK.  Then  will  the  bases  AB  and  AH 
be  commensurable,  and  ABCD :  AHKD : :  AB ;  AH.  But, 
by   supposition,  ABCD  :  AFGD ::  AB  :  AL.      Since  the 


BOOK   V. 


97' 


antecedents  in  both  proportions  are  the  same,  the  con- 
sequents form  a  new  proportion,  AHKD :  AFGD  ::  AH : 
AL.  But  AL  is  greater  than  AH ;  therefore  AFGD  is 
greater  than  AHKD,  which  is  absurd.  Therefore  ABCD 
can  not  be  to  AFGD  as  AB  is  to  a  line  greater  than  AP. 
And,  similarly,  it  can  be  proved  that  the  fourth  propor- 
tional can  not  be  less  than  AF.  Hence,  ABCD :  AFGD 
::AB:AF. 


Pkoposition  n. — Theorem. 

Any  two  rectangles  are  to  each  other  as  the  product  of 
their  bases  and  altitudes. 

Let  ABDC  and  AGFH  Ef 
be    two    rectangles;    then 
will  ABDC  be  to  AGFH  as 
ABxAC:AGxAH. 

Place  the  two  rectangles 
so  that  the  angles  at  A  may 
be   vertical;    produce    DC 
and  FH  until  they  meet  at  j« 
E. 

The  two  rectangles  ABDC  and  ACEH,  having  the  same 
altitude,  AC,  are  to  each  other  as  their  bases,  AB  and  AH. 
In  like  manner,  the  two  rectangles  AGFH  and  ACEH, 
having  the  same  altitude,  AH,  are  to  each  other  as  their 
bases,  AG  and  AC.     Hence  there  are  two  proportions : 

ABDC :  ACEH  ::AB:  AH, 
and        ACEH :  AGFH : :  AC :  AG. 

By  multiplying  these  two  proportions,  observing  cO 
strike  out  ACEH,  which  appears  both  in  the  antecedent 
and  consequent,  there  will  be  a  new  proportion ; 
ABDC :  AGFH : :  AB  x  AC :  AH  x  AG. 

Scholium,  Suppose  AB  divided  into  11  equal  parts, 
and  AC  into  4  equal  parts,  and  suppose  the  unit  of  meas- 
ure be  one  yard ;  then  the  area  of  the  rectangle  ABCD 
w^ill  be  44  square  yards;  that  is,  11  x4r=44  square  yards. 
By  actual  count,  there  will  be  found  44  little  squares, 

E 


98 


ELEMENTS  OP  PLANE  GEOMETRY. 


4 

3 

2 

1 

£ 

3 

4 

5 

6 

y 

8 

9 

iO 

11 

A 

B 

1)  each  of  which  is  a 
square  yard.  Hence, 
to  find  the  area  of  a 
rectangle  or  of  a  parol- 
lelogram^  multiply  the 
base  by  the  altitude; 
and,  to  find  the  area 
of  a  triangle,  multiply  the  base  by  one-half  the  altitude, 
or  the  altitude  by  one-half  the  base  ;  or  multiply  the  base 
by  the  altitude  and  take  half  the  product. 

Cor,  Since  a  triangle  is  half,  a  rec^an^gle  pr  parallelo- 
gram, having  the  same  base  afitd  aiti:tude ;  therefore  tri- 
angles on  the  same  base  and  between  the  same  parallels 
are  to  each  other  as  their  bases. 


Proposition  III.— Theorem. 

A  line  drai07i. parallel  to  one  side  of  a  triangle  divides 
the  other  tico  sides  hito  proportional 2yarts, 

C  Let  ABC  be  any  triangle,  and  DE  a 

line  drawn  parallel  to  AB ;  then  will 
DE  divide  AC  and  BC  into  propor- 
tional parts;  that  is,  DC:AD::CE: 
EB. 

For,  draw  the  lines  AE  and  BD. 
The  triangles  ADE  and  BDE  are  equal, 
■^  because  they  have  the  same  base,  DE, 
and  are  between  the  same  parallels, 
DE  and  AB  (Prop.  VIIL,  Bk.  II.).  But 
the  triangles  ADE  and  DEC  have  their  bases  in  the 
same  line,  AC,  and  have  the  same  altitude.    Therefore 

ADE :  CDE : :  AD :  DC  (Prop.  II.,  Cor.) ; 
and,  for  the  same  reason, 

BDE:  CDE::  BE:  EC. 
And,  since  ADE  is  equal  to  BDE,  and  CDE  common  in 
both  proportions,  by  equality  of  ratios,  AD: DC:: BE: 
EC. 


BOOK  V. 


99 


Cor.  1.  By  composition  (Prop.  VII.,  Bk.  IV.),  AD+ 
DC:DC::BE+EC:EC,  or  AC  :  DC::BC:EC,  or  AC: 
AD::BC:BE. 

Cor,  2.  If  any  number  of  lines  be  drawn  parallel  to  a 
side  of  a  triangle,  each  parallel  cuts  the  other  two  sides 
proportionally. 

Proposition  IV. — Theorem. 

If  a  straight  U?ie  cut  two  sides  of  a  triangle  proportion- 
ally^  it  will  be  parallel  to  the  other  side. 

Let  the  straight  line  DE  cut  the 
two  sides  of  the  triangle  ABC  pro- 
portionally ;  then  will  DE  be  par- 
allel to  AB. 

Draw  AD  and  BE.  The  tri- 
angle ADE :  DEC  : :  AE :  EC  (Prop. 
II.,  Cor.),  andBDE:DEC::BD: 
DC.  But,  by  hypothesis,  AE  :  EC 
: :  BD :  DC.  Hence,  by  equality  of 
ratios,  ADE :  DEC : :  BDE :  DEC. 
have  the  same  ratio  to  DEC.  Hence  they  are  equal. 
Now  if  the  triangles  ADE  and  BDE  are  equal,  and  have 
the  same  base,  AB,  it  is  evident  that  they  must  have  the 
same  altitude ;  that  is,  AB  and  ED  are  parallel. 

Proposition  Y. — Problem. 
To  divide  a  line  into  any  member  of  equal  parts. 

Let  AB  be  the 
given  line :  it  is  re- 
quired to  divide  it 
into  any  number  of 
equal  parts,  say  five. 

Draw  the  indefinite 

line  AM,  making  any 

angle  with  AB.     On  ^       ^      D     B     F     D 

AM  cut  off  five  equal  parts,  A«,  ccy,  yz^  »r,  and  rru    Join 


But  ADE  and  BDE 


100 


ELEMENTS  OF  PLANE  GEOMETRY. 


B  and  n.  Then,  through  the  points  cc,  y^  z^  r,  draw  the 
lines  Qx^  D^,  E^,  and  Yr  parallel  to  Bn.  These  parallel 
lines  divide  AB  and  An  proportionally  (Prop.  Ill,  Cor. 
2).  But  An  is  divided  into  five  equal  parts  (const.). 
Therefore  AB  is  divided  in  five  equal  parts. 


Proposition  VT. — Problem. 


To  find  a  third  proportional  to  two  given  lines. 

Let    a    and    b   be 
G  ^^ 


a  and  b  be  two 
given  lines :  it  is  required 
to  find  a  third  propor- 
tional. 

Draw  CD  and  CM,  mak- 
ing any  angle.  On  CD  and  CM  cut  off  parts  CF  and 
CG  equal  to  a,  and  on  CD  cut  off  a  part  CE  equal  to  b. 
Join  E  and  G,  and  through  the  point  F  draw  FM  paral- 
lel to  EG.  CM  is  the  third  proportional.  For  CE:CF 
::  CG :  CM  (Prop.  III.).  But  CF  and  CG  are  each  equal 
to  a,  and  CE  is  equal  to  b.    Therefore  b:a\ : a :  CM. 


Proposition  VII. — Problem. 

To  firid  a  fourth  proportional  to  three  given  lines. 

Let  a,  ^,  and  c  be 
three  given  lines :  it  is 
required  to  find  a  fourth 
proportional. 

Draw  two  lines  AC 
and  AB,  making  any 
angle.  On  AC  cut  off 
AD  equal  to  a,  and  AE 
equal  to  b  ;  on  AB  cut  off  AG  equal  to  c.  Join  D  and  G, 
and  through  the  point  E  draw  EF  parallel  to  DG.  AF 
is  the  fourth  proportional.  For  AD :  AE : :  AG :  AF ;  but 
AD  is  equal  to  a,  AE  to  b,  and  AG  to  c.  Hence  a\by, 
c:AF. 


BOOK  V.  lo; 

Proposition  YIII. — Theorem. 

Equiangular  triangles  are  similar^  or  have  their  homol- 
ogous sides  proportio7iaL 

Let  the  two 
triangles 
ABC  and 
DEF  be  equi- 
angular,    the  

angle  A  equal  ^  ^    ^  Sr  53 

to  the  angle  D,  the  angle  B  equal  to  the  angle  E,  and  the 
angle  C  equal  to  the  angle  F;  then  will  these  two  trian- 
gles be  similar,  and  AB  w^ill  be  to  AC  as  DE  to  DF,  and 
so  on. 

For  from  DE  cut  off  the  part  DH  equal  to  AB,  and 
from  DF  cut  off  the  part  DG  equal  to  AC.  Join  G  and 
H.  The  two  triangles  ABC  and  DHG  have  two  sides  of 
the  one  equal  to  two  sides  of  the  other  (const.),  and  the 
included  angles  A  and  D  equal  (hyp.) ;  therefore  they  are 
equal  in  all  their  parts  (Prop.  IV.,  Bk.  I.),  and  the  angle 
DHG  is  equal  to  the  angle  B ;  but  the  angle  E  is  also 
equal  to  the  angle  B.  Therefore  the  angle  DHG  is  equal 
to  the  angle  E  (Ax.  1) ;  and  since  DHG  is  equal  to  E, 
HG  must^be  parallel  to  EF  (Prop.  XXVI.,  Bk.  I.).  Be- 
cause HG  is  parallel  to  EF,  DH:  DE  ::DG:DF  (Prop. 
HI.).  By  substituting  for  DH  and  DG  their  equals,  AB 
and  AC,  then  AB :  AC : :  DE :  DF. 

Proposition  IX. — Theorem. 

Two  triangles  which  have  their  homologous  sides  jy^'o- 
portio7ial  are  equiangidar  and  similar. 

Let  the  two  triangles  ABC  and  DEF  have  their  homol- 
ogous sides  proportional,  DF :  DE  : :  AC  :  AB,  and  DE :  EF 
: :  AB  :  BC ;  then  will  the  two  triangles  be  equiangular 
and  similar. 

For  at  the  point  D  make  the  angle  GDE  equal  to  A, 


1 02 


KIEiJENTS    OP   PLANE   GEOMETRY. 


and  at  the  point  E  make 
the  angle  DEG  equal  to 
B;  and  since  the  sum  of 
the  angles  GDE  and  DEG 
is  less  than  two  right  an- 
gles, the  lines  DG  and  EG 
must  meet,  and  the  angle 

G  must  equal  the   angle 

^  B  C  (Prop.  XXX.,  Cor.,Bk. 

I.).     Hence  the  triangles  ABC  and  DGE  are  equiangular, 
and  AB:BC::DE:EG 

(by  previous  Prop.),  but,  by  hypothesis, 

AB:BC::DE^:EF; 
since  the  three  first  terms  are  identical,  it  follows  that 
EG  is  equal  to  EF.     Again, 

AB:AC::ED:DG 
(by  previous  Prop.),  but,  by  hypothesis, 

AB:AC::ED:DF; 
since  the  three  first  terms  are  identical,  DG  is  equal  to 
DF.  The  two  triangles  DEG  and  DEF  have  their  three 
sides  equal,  each  to  each ;  they  are,  therefore,  equal  in 
all  their  parts,  and  quiangular  (Prop.  VII.,  Bk.  I.).  But 
ABC  is  equiangular  to  DEG.  Hence  ABC  is  also  equi- 
angular to  DEF. 

Pkoposition  X. — Theorem. 

If  two  triangles  have  an  angle  in  each  equals  and  the 
sides  about  these  equal  angles  proportional^  they  icill  be 
equiangular. 

Let  the  two  triangles 
ABC  and  DEF  have 
the  angle  A  equal  to 
the  angle  D,  and  the 
side  AB  to  the  side 
AC  as  DE  to  DF; 
then  will  these  trian- 
gles be  equiangular. 


BOOK  y.  103 

For  from  DE  cut  off  the  part  DII  equal  to  AB,  and 
from  DF  cut  off  the  part  DG  equal  to  AC.  Join  G  and 
H.  The  two  triangles  ABC  and  DHG  have  two  sides 
of  the  one  equal  to  two  sides  of  the  other,  Qach  to  each 
(const.),  and  the  included  angles  A  and  D  equal.  They 
are,  therefore,  equal  in  all  their  parts  (Prop.  IV.,  Bk.  I.) ; 
the  angle  DGH  is  equal  to  C,  and  DHG  is  equal  to  B. 

By  hypothesis, 

AB:AC::DE:DF; 
but  DH  and  DG  are  equal  to  AB  and  AC  (const.).    There- 
fore 

DE:DF::DH:DG. 
Hence  GH  is  parallel  to  FE  (Prop.  IV.).  Consequently 
the  angle  F  is  equal  to  the  angle  DGH,  and  the  angle  E 
is  equal  to  the  angle  DHG  (Prop.  XXVIL,  Bk.  I).  But 
the  angles  C  and  B  were  before  proved  equal  to  DGH 
and  DHG.  Therefore  the  angles  C  and  B  are  equal  to 
the  angles  F  and  E,  each  to  each,  and  the  triangles  ABC 
and  DEF  are  equiangular. 

Proposition  XI. — Theorem. 

A  Ihie  which  bisects  any  angle  of  a  triangle^  divides  the 
ojyposite  side  into  seginents  proportional  to  the  other  txco 
sides. 

Let  the  line  DB  bisect 
the  angle  ABC  of  the 
given  triangle  ACB ; 
then  will  the  segments 
AD  and  DC  be  propor- 
tional to  the  other  two 
sides  AB  and  BC,  or  AD 
:DC::AB:BC.  ^ 

For,  from  the  point  C 
draw  CE  parallel  to  DB  and  produce  AB,  until  it  meets 
CE  at  the  point  E.  Since  DB  and  CE  are  parallel,  the 
angle  BCE  is  equal  to  the  angle  DBC  (being  alternate 
angles) ;  and  the  angle  E  is  equal  to  the  angle  ABD  [an 


104 


ELEMENTS  OF  PLANE  GEOMETRY. 


exterior  angle  equal  to  an  interior  and  opposite  angle  on 
the  same  side  of  the  secant  line]  (Prop.  XXVII.,  Bk.  I.)  ; 
and  since  the  angles  E  and  BCE  are  each  equal  to  half 
of  the  angle  ABO,  they  are  equal  to  each  other,  and  the 
triangle  BCE  is  isosceles  (Prop.  XIX.,  Bk.  I.).  Hence 
CB  is  equal  to  BE.  Since  DB  is  parallel  to  CE,  AD :  DC 
::  AB  :  BE  (Prop.  III.),  but  BC  is  equal  to  BE.  There- 
fore AD :  DC ::  AB :  BC. 


Proposition  XII. — ^Theorem. 

Two  triangles  lohich  have  their  homologous  sides  jjaral- 
lei  or  perpendicular^  are  similar. 

1.  Let  the  two  triangles  ABC 
and  DEF  have  AB  parallel  to 
DE,  AC  parallel  to  DF,  and  BC 
parallel  to  EF ;  then  will  they 
be  similar. 

For,  since  AC   and  CB   are 

parallel  to  DF  and  FE,  each  to 

each, ^  the  angles  C  and  F  are 

equal  (Prop.  ^ XXXIII.,  Bk.  I.). 

Similarly  it  can' be  proved  that 

-^  I^^    the  angles  A  and  D,  and  B  and 

E  are  equal.     Hence  the  triangles  are  equiangular  and 

similar  (Prop.  VIII.). 

2.  Let  the  triangle 
DEF  have  its  three 
sides  DE,  EF,  and 
FD,  perpendicular  re- 
spectively to  CB,  CA, 
and  AB;  then  will 
these  two  triangles  be 
similar. 

A  KB     YoY,    prolong     DE, 

FE,  and  FD,  until  they  meet  the  sides  in  H,  G,  and  K. 
In  the  quadrilateral  AKFG,  the  angles  AKF  and  AGF 


BOOK  V.  105 

are  right  angles ;  hence  the  sum  of  the  angles  A  and 
GFK  is  equal  to  two  right  angles  (Prop.  XXX.,  Bk.  I., 
Cor.  1.).  But  the  sum  of  the  angles  DFE  and  GFK  is 
equal  to  two  right  angles  (Prop.  XV.,  Bk.  I.).  Take 
away  the  common  angle  GFK  from  these  two  equations, 
and  there  will  remain  the  angle  A  equal  to  the  angle 
DFE ;  and  similarly  it  can  he  proved  that  the  angle  C  is 
equal  to  the  angle  DEF,  and  that  the  angle  B  is  equal  to 
the  angle  EDF.  Hence  the  two  triangles  are  equiangu- 
lar and  similar. 

Proposition  XIII. — Theorem. 

Two  triangles  having  an  angle  in  each  equal ^  are  to 
each  other  as  the  rectangles  of  the  sides  which  co7iiain  the 
equal  angles. 

Let  the  two  tri-  T  /\C 

angles  ABC  and 
DEF  have  the  an- 
gles C  and  F  equal ; 
then  will  ABC  be 
to  DEF  as  AC  x 
CBiDFxFE. 

For,  make  CG 
equal  to  DF,  and 
CH  equal  to  FE.  A^ 

The  triangles  DEF  and  CGH  are  equal  (Prop.  IV.,  Bk.  I), 
and  in  all  respects  identical.  Join  G  and  B.  The  trian- 
gles ABC  and  BCG  have  the  same  altitude  and  their 
bases  in  the  same  line.  Hence  the  triangle  ABC :  the 
triangle  CBG ::  AC :  GC  (Prop.  II.,  Cor.),  and  the  triangle 
CBG :  the  triangle  CGH : :  CB :  HC  for  the  same  reason. 
By  multiplying  these  two  proportions,  and  striking  out 
CBG,  which  is  both  an  antecedent  and  a  consequent,  a 
new  proportion  is  obtained ;  ABC :  CGH : :  AC  x  CB :  GC 
X  HC,  or,  substituting  for  GC  and  CH  their  equals  DF 
and  FE,  and  for  the  triangle  CGH  its  equal  DEF,  ABC : 
DEF ::  AC  xCB :  DFxFE. 

E2 


1Q6         ELEMENTS  OF  PLANE  GEOMETRY. 


Proposition  XIY. — Thjeorem. 

If  a  perpendicular  he  drawn  from  the  right  angle  of  a 
right-angled  triangle  to  the  hypothenuse^  it  divides  the  tri- 
angle  into  segments  similar  to  the  whole  triangle  and  to 
each  other :  the  perpendicidar  is  a  mean  proportional  be- 
tween the  segments  of  the  hypothenuse^  and  either  side  is 
a  mean  proportional  between  the  hypothenuse  and  the  ad- 
jacent segmeyit, 

1.  Let  ABC  be  the  right- 
angled  triangle,  and  CD  the 
perpendicular ;  then  will  the 
triangle  ADC  be  similar  to 

^^ •  the  triangle  ABC  ;  DBC  will 

A  I)      U   also  be  similar  to  ABC,  and 

ADC  and  DBC  will  be  similar  to  each  other. 

For,  in  the  two  triangles  ABC  and  ADC,  the  angles 
ACB  and  ADC  are  equal  (each  being  a  right  angle),  the 
angle  A  is  common.  Hence  the  remaining  angles  ABC 
and  ACD  are  equal  (Prop.  XXX.,  Bk.  I.,  Cor.),  and  the 
triangles  are  similar.  In  like  manner  it  can  be  proved 
that  ABC  and  DBC  are  also  similar.  In  the  two  trian- 
gles ADC  and  DBC,  the  angles  at  D  are  right  angles, 
and  the  angle  B  has  just  been  shown  equal  to  ACD,  and 
A  equal  to  DCB.  Hence  the  triangles  ADC  and  DBC 
are  similar. 

2.  DC  will  be  a  mean  proportional  between  AD  and 
DB.     For,  since  the  triangles  ADC  and  DBC  are  similar., 

AD:DC::DC:DB; 
and,  since  ABC  and  ADC  are  similar, 

AB:AC::AC:AD; 
and,  in  like  manner, 

AB:CB::CB:DB. 
Cor,  1.  DC^rrrADxDB;  AC^^ABxAD,  and  CB2== 
ABxDB  (Prop.  II.,  Bk.  IV.). 


BOOK   V. 


107 


Proposition  XV. — Problem. 

To  find  a  mean  proportional  between  two  given  lines. 

Let  A  and  B  be  the  two  given      A 

lines:  it  is  required  to  find  a 
mean  proportional  between  them. 
Place  the  two  lines  A  and  B  so 
that  they  will  form  one  continued 
line  DE,  with  DF  equal  to  A, 
and  FE  equal  to  B.  Describe 
the  semicircle  of  which  DE  is 
the  diameter.  At  the  point  F  erect  the  perpendicular 
FC.  Join  D  and  C,  and  E  and  C.  The  triangle  DCE 
is  a  right-angled  triangle,  because  DCE  is  an  angle  in- 
scribed in  a  semicircle.     Hence  DF :  CF : :  CF :  FE. 

Proposition  XVI. — Problem. 

To  construct  a  rectangle  equivalent  to  a  giveii  square^  and 
having  th^  sum  of  its  adjacent  sides  equal  to  a  given  line. 

Let  S  be  the  given 
square,  and  AB  equal 
to  the  sum  of  the 
sides  of  the  rect- 
anocle. 

Upon  the  diam-  ^ 
eter  AB  describe  the  semicircle  ACDB,  and  draw  CD 
parallel  to  AB,  CD  being  distant  from  AB  the  side  of 
the  given  square ;  then  from  the  point  D  draw  the  per- 
pendicular DE.  AE  and  EB  will  be  the  sides  of  the  re- 
quired rectangle. 

For  DE  is  a  mean  proportional  between  4^E  and  EB ; 
that  is  (Prop.  XIV.,  Cor.),  DE^^AExEB. 

Proposition  XVTT. — Theorem. 
Similar  triangles  are  to  each  other  as  the  squares  of 
their  homologous  sides. 

Let  the  two  triangles  ABC  and  DEF  be  similar,  the 


108 


ELEMENTS  OF  PLANE  GEOMETRY. 


A. 


angle  A  equal  to  the  angle  D,  the 
angle  B  equal  to  the  angle  E,  and 
the  angle  C  equal  to  the  angle 
F;  then  will  the  triangles  be  to 
each  other  as  the  squares  of  their 
homologous  sides. 

For  the  triangle  ABC :  the  tri- 
DFxFE  (Prop.XIIL),  and  AC: 


angle  DEF ::  AC  xCB: 

DF : :  CB :  FE  (Prop.  VIII.) ;  multiply  the  terms  of  this 
proportion  by  the  identical  proportion, 
CB:FE::CB:FE/ 
and  the  result  is 

ACxCB:DFxFE::CB2:FE2. 
By  substituting  for  the  two  first  terms  their  equivalent 
ratio,  the  triangle  ABC:  the  triangle  DEF : :  CB^ : FE2, 
and  in  like  manner  it  can  be  proved  of  any  other  two 
homologous  sides. 

Pkoposition  XVni. — Theorem. 

If  from  a  point  without  a  circle  a  tangent  and  a  secant 
be  draion,  the  tangent  will  he  a  mean  proportional  between 
the  secant  and  its  external  segment. 

Let  AC  be  a  secant,  and  CB  a 
tangent,  both  drawn-  from  the  point 
C ;  then  will  CB^  be  equal  to  AC  x 
DC. 

For,  joining  the  points  A  and  B, 
and  B  and  D,  the  triangles  ABC 
and  CBD  will  have  the  angle  C 
common,  and  the  angle  CBD  equal 
to  the  angle  A  (Prop.  XY.,  Bk.  III.). 
Hence  the  remaining  angles  ABC 
and  CDB  are  equal,  and  the  two  triangles  are  equiangu- 
lar and  similar. 

Therefore  AC :  CB : :  CB : :  DC, 

or  CB2=ACxDC. 


BOOK  V. 


109 


Propositiox  XIX. — Theorem. 

If  two  chords  intersect  each  other ^  they  are  reciprocally 
proportional. 

Let  the  two  chords  AB  and  CD  inter- 
sect each  other ;  then  will  AO  be  to  DO 
as  CO  to  BO. 

For,  joining  A  and  C,  and  B  and  D, 
the  two  triangles  AOC  and  BOD  are 
equiangular,  because  the  angles  AOC 
and  BOD  are  equal,  being  vertical,  and 
A  is  equal  to  D,  and  C  is  equal  to  B  (Prop.  X.,  Cor.  5, 
Bk.III.).     Hence  AO:  DO::  CO:  OB. 

Cor.  By  making  the  product  of  the  extremes  equal  to 
the  product  of  the  means,  AOxOB=DOxOC;  that  is, 
the  rectangle  of  the  two  segments  of  one  chord  is  equal 
to  the  rectangle  of  the  two  segments  of  the  other  chord. 


Proposition  XX. — Theorem. 

If  from  a  point  without  a  circle  tioo  secants  he  drawn 
terminating  in  the  concave  arc^  the  whole  secants  will  he 
reciprocally  proportional  to  their  external  segments. 

Let  AB  and  AC  be  two  secants  ter-  A 

minating  in  the  concave  arc;  then  will 
AB  be  to  AC  as  AE  to  AD. 

For,  drawing  BE  and  DC,  the  triangles 
ABE  and  ACD  will  have  the  angle  A 
common,  and  the  angle  at  B  equal  to  the 
angle  at  C,  because  each  is  measured  by- 
half  the  arc  DE  (Prop.^X.,  Cor.  5,  Bk. 
IIL).  Hence  these  triangles  are  equi- 
angular and  similar.     Therefore,  AB :  AC  : :  AE :  AD. 

Cor,  By  making  the  product  of  the  extremes  equal  to 
the  product  of  the  means, 

ABxAD^ACxAE. 


110 


ELEMENTS  OF  PLANE  GEOMETRY. 


Pkoposition  XXI. — Theorem. 

In  every  triangle  the  rectangle  contained  by  any  two 
sides  is  equal  to  the  rectangle  contained  by  the  diameter 
of  the  circumscribing  circle  and  the  perpendicular  drawn 
to  the  third  side  from  the  opposite  angle. 

Let  ABC  be  the  given  triangle,  CE 
the  perpendicular  on  AB,  and  BD  the 
diameter  of  the  circle  ABCD;  then 
willACxCBrrrDBxCE. 

For,  drawing  DC,  the  triangles  DCB 
and  ACE  are  right  angled,  ACB  being 
inscribed  in  a  semicircle,  and  AEC  by- 
construction,  and  the  angle  A  is  equal  to  the  angle  I) 
(Prop.  X,  Cor.  5,  Bk.  III.).  Hence  the  triangles  are  equi- 
angular and  similar ; 
and  AC:DB::EC:CB; 

therefore  AC  x  CB=DB  x  EC. 


Proposition  XXII. — Theorem. 

If  a  line  be  drawn  bisecting  any  angle  of  a  triangle 
and  terminating  in  the  opposite  side^  the  rectangle  con- 
tained by  the  sides  of  this  angle  will  be  equal  to  the  rect- 
angle of  the  segments  of  the  third  side,  together  with  the 
square  of  the  bisecting  line. 

Let  BD  bisect  the  angle  ABC ;  then 
will  ABxBC=ADxDC+BD2. 

Describe  a  circumference  which 
shall  pass  through  the  points  A,  B, 
and  C  (Prop.  Y.,  Bk.  IIL),  produce  BD 
to  E,  and  join  EC. 

In  the  two  triangles  ADB  and 
BCE,  the  angles  ABD  and  CBE  are 
equal  (const.),  and  the  angle  E  is  equal  to  the  angle  A 
(Prop.  X.,  Cor.  5,  Bk.  HI.).  Hence  the  two  triangles  are 
similar.     AB :  BE : ;  BD :  BC ;  making  the  product  of  the 


BOOK   V.  Ill 

extremes  equal  to  the  product  of  the  means,  BAxBC= 
BExBD;  by  substituting  BD+DE  for  BE,  then  ABx 
BCz:zBDx(BD  +  DE);  that  is,  ABxBC=:BD2+BDx 
DE;  but  BDxDE  is  equal  to  ADxDC  (Prop.  XIX.); 
and,  by  substitution,  AB  x  BC=:  AD  xDC+BD2. 

Proposition  XXIII. — Theorem. 

Two  similar  polygoyis  are  composed  of  the  same  num- 
ber of  triangles^  similar  to  each  other ^  and  similarly  sit- 
uated. 

Let  ABODE  and 
FGKLM  be  two  sim- 
ilar polygons ;  from 
any  angle,  A,  draw  AD 
and  AC,  and  from  the 
angle  F,  homologous  •^z' 
to  A  in  the  other  poly- 
gon, draw  FL  and  FK. 

Since  the  polygons  -^ 
are  similar,  the  angle 
E  must  be  equal  to  the  angle  M;  and  the  sides  which 
contain  these  equal  angles  are  proportional,  AE :  ED : : 
FM :  ML.  Therefore  the  two  triangles  AED  and  FML, 
having  an  angle  in  each  equal,  and  the  sides  containing 
the  angles  proportional,  are  similar  (Prop.  X.),  and,  since 
they  are  similar,  the  angle  EDA  is  equal  to  the  angle 
MLF ;  but  the  angle  EDO  is  equal  to  the  angle  MLK, 
and  if  the  former  equals  be  taken  from  the  latter  equals, 
ADC  will  remain  equal  to  FLK.  Since  the  triangles  are 
similar,  AD  is  proportional  to  FL ;  and,  since  the  poly- 
gons are  similar,  DC  is  proportional  to  LK.  The  two 
triangles  ADC  and  FLK  have  the  angles  ADC  and  FLK 
equal,  and  the  sides  which  contain  these  angles  propor- 
tional. Therefore  they  are  similar  (Prop.  X.).  In  the 
same  manner  it  can  be  proved  that  ABC  is  similar  to 
FGK. 

Scholiimi,  The  converse  of  this  proposition  is  also  true. 


112         ELEMENTS  OF  PLANE  GEOMETRY. 

If  two  polygons  are  composed  of  the  same  number  of 
similar  triangles  similarly  situated,  these  polygons  will 
be  similar. 

For  the  similarity  of  the  respective  triangles  will  give 
the  angle  E  — to  M,  BrrG,  and,  by  adding  the  equal  an- 
gles at  A  and  F,  EAB  will  be  equal  to  MFG,  and,  in  like 
manner,  EDO  and  DCB^MLK  and  LKG. 

Proposition  XXIY. — Theorem. 

The  perimeters  of  shnilar  polygons  are  to  each  other  as 
their  homologous  sides  ;  and  their  areas  are  to  each  other 
as  the  squares  of  those  sides, 

1.  Since  the  polygons 
are  similar,  AB :  FG : : 
BC:GK::DC:LK,  etc. 
Now,  as  the  sum  of  all 
the  antecedents  is  to 
the  sum  of  all  the  conse- 
quents as  any  one  ante- 

cedent  is  to  any  one  con- 

^  ^  ^  sequent,   AB+BG+DC 

-fED  +  AE:FG  +  GK  +  KL  +  LH  +  FH::AB:FG;  or, 
since  the  sum  of  these  antecedents  and  of  these  conse- 
quents will  be  the  perimeters  of  the  polygons, 

ABODE :  FGKLH ::  AB :  FG, 
and  so  of  any  other  two  homologous  sides. 

2.  Since  the  triangles  ABC  and  FGK  are  similar,  the 
triangle  ABC:  the  triangle  FGK::AC2:FK2  (Prop. 
XYIL),  and,  for  a  like  reason,  the  triangle  ACD :  the  tri- 
angle FLK : :  AC^  :  FK^.  But  in  these  two  proportions 
the  last  couplets  are  identical ;  hence  the  triangle  ABC : 
FGK::  ACD:  FLK.  In  like  manner  it  can  be  proved 
that  ACD:FLK::AED:FHL,  and  so  on.  By  adding 
the  antecedents  and  consequents  ABC-|-ACD+AED: 
FGK+FLK+FHL::ABC:FGK;  that  is,  the  polygon 
ABCDE :  FGKLH : :  ABC :  FGK.    But  these  triancrles  are 


Au 


BOOK  V.  113 

to  each  other  as  the  squares  of  their  homologous  sides. 
Hence  ABCDE :  FGKLII : :  AB2 :  FG2. 

Cor.  If  three  similar  figures  are  constructed  on  the 
three  sides  of  a  right-angled  triangle,  the  figure  on  the 
hypothenuse  will  be  equal  to  the  sum  of  the  other  two ; 
for  the  three  figures  are  to  each  other  as  the  squares  of 
their  homologous  sides.  But  the  square  of  the  hypoth- 
enuse is  equal  to  the  sum  of  the  squares  of  the  other  two 
sides.  Hence  the  similar  figure  descijibed  on  the  hypoth- 
enuse is  equal  to  the  sum  of  the  two  similar  figures  de- 
scribed on  the  other  two  sides. 

Pkoposition  XXV. — Theorem. 

If  a  quadrilateral  he  inscribed  in  a  circle^  the  rectangle, 
of  the  two  diagonals  is  equal  to  the  sum  of  the  rectangles 
of  the  oj^posite  sides,  taken  two  by  two. 

Let  ABCD  be  a  quadrilateral  in-  jD 

scribed  in  a  circle,  and  AC  and  BD  X^^ 
its  diagonals ;  then  will  AC  x  BD  be  f  ^y^ 
equal  to  AD  x  BC  plus  AB  x  DC.         -^(^^^ 

For,  making  the  arc  HC  equal  to  \\  / 
AB,  and  joining  HD,  the  triangles  ^/ 
ADB  and  GDC  are  equiangular  and  ^C 
similar,  because  the  angles  ADB  and 
GDC  are  equal  (Prop.  X.,  Cor.  5,  Bk.  HI),  each  being 
measured  by  the  half  of  equal  arcs ;  and  the  angle  ABD 
is  equal  to  the  angle  GCD,  because  each  is  measured  by 
half  the  arc  AD.  Hence  the  two  remaining  angles  BAD 
and  CGD  are  equal.     Therefore, 

AB:GC::DB:DC, 
and,  making  the  rectangle  of  the  extremes  equal  to  the 
rectangle  of  the  means, 

ABxDC=GCxDB. 

Again,  the  two  triangles  BDC  and  AGD  are  equiangu- 
lar and  similar,  because  the  angle  BDC  is  equal  to  the 
angle  ADG,  each  being  measured  by  the  half  of  equal 
arcs  BC  and  AH,  for  to  the  equal  arcs  AB  and  HC,  the 


114 


ELEMENTS    OF   PLANE    GEOMETRY. 


arc  BH  is  added,  and  the  angle  DBC  is  equal  to  the  angle 
DAG,  each  being  measured  by  half  the  arc  DC.  Hence 
the  two  remaining  angles  BCD  and  AGD  are  equal. 
Therefore 

AD:DB::AG:BC; 
and  hence  AD  x  BC =DB  x  AG. 

By  adding  the  two  equations, 

ABxDC+ADxBC=GCxDB+AGxDB; 
ABxDC+ADxBC=:(GC+AG)xDB,or=ACxDB. 


The  folloicing  are  Test  Examples  involving  Book 
Fifth: 

1,  To  bisect  a  quadrilateral  by  a  line  drawn  from  one 
of  its  angles. 

Let  DCB  be  the  an- 
gle from  which  the  line 
shall  be  drawn  bisect- 
ing the  given  quadri- 
lateral ABCD. 

Draw  the  diagonals 
DB  and  AC;  bisect 
DB  in  the  point  E,  and 
through  E  draw  FEG 
parallel  to  AC.  Join  AE  and  EC,  and  from  C  draw  CF. 
CF  bisects  the  quadrilateral. 

The  triangles  DEC  and  BEC  are  equal  (Prop.  VIII., 
Bk.  II.),  and  also  AED  and  AEB,  for  the  same  reason. 
By  adding  AED  and  EDC,  and  AEB  and  ECB,  two  by 
two,  the  quadrilateral  AECD  will  be  equal  to  the  quadri- 
lateral AECB.  The  triangles  AFE  and  FEC  are  equal 
(Prop.  VIII.,  Bk.  II.).  From  each  take  away  the  common 
triangle  FHE,  and  there  will  remain  AFH  equal  to  HEC. 
Now  if  from  the  quadrilateral  AECD  the  triangle  HEC 
be  taken  aw^ay,  and  its  equal  AFH  added,  we  shall  have 
the  quadrilateral  AFCD ;  and  if  from  AECB  AFH  be. 
taken  away,  and  its  equal  HEC  added,  we  shall  have  the 


BOOK   V.  115 

triangle  FOB.     Hence  FCB  is  equal  to  AFCD,  and  FC 
bisects  the  quadrilateral. 

2.  To  determine  the  figure  formed  by  joining  the  points 
of  bisection  of  the  sides  of  a  trapezimuy  aiidits  ratio  to  the 
trapezium. 

Let  ABCD  be  the 
given  trapezium,  and  let 
the  sides  AB,  CB,  DC, 
and  AD  be  bisected  in 
the  points  E,  H,  F,  and 
G.  By  joining  these 
points  of  bisection,  a  par- 
allelogram, GEHF,  is 
formed.  B 

For,  since  AD  and  DC  are  bisected  in  G  and  F,  AD : 
GD : :  DC :  DF,  and  hence  GF  is  parallel  to  AC ;  and  in 
the  same  manner  EH  is  parallel  to  AC.  Therefore  GF 
and  EH  are  parallel  to  each  other.  In  like  manner,  it  can 
be  proved  that  FH  and  GE  are  parallel  to  each  other. 
Hence  GEHF  is  a  parallelogram. 

Again,  the  triangle  ADC  is  composed  of  the  triangles 
ALG,GND,DNF,FKC  (without  the  parallelogram),  and 
of  the  triangles  GLN,  LNO,  ONK  and  NFK  (within  the 
parallelogram).  Now  it  is  evident  that  ALG  and  GND 
are  equal  to  LNG  and  LNO,  and  also  that  DNF  and 
FKC  are  equal  to  NFK  and  ONK.  By  adding  these,  it 
will  be  found  that  the  parallelogram  GLKF  is  equal  to 
the  sum  of  the  triangles  ALG,  GND,  NDF,  and  FKC. 
Li  the  same  manner,  it  can  be  proved  that  the  parallelo- 
gram LEHK  is  equal  to  the  sum  of  the  triangles  ALE, 
EMB,  MBH,  and  HKC.  Hence  the  whole  parallelogram 
is  one  half  the  trapezium. 

3.  If  from  any  point  in  the  base  of  an  isosceles  triangle 
perpendiculars  be  drawn  to  the  sides,  these  together  shall 
be  equal  to  a  perpendicular  drawn  from  either  extremity 
of  the  base  to  the  opposite  side. 

Let  ABC  be  an  isosceles  triangle ;  and  from  any  point, 


116  ELEMENTS    OF   PLANE    GEOMETRY. 

D,  draw  the  perpendiculars  DE  and 
DF;  also  the  perpendicular  BG.  It 
is  required  to  prove  that  BG  is  equal 
to  the  sum  of  DE  and  DF. 

In  the  two  triangles  BED  and 
DCF,  the  angles  EBD  and  C  are 
equal  (hyp.),  and  BED  and  DFC  are 
also  equal,  being  right  angles.  Hence 
the  triangles  are  similar,  and  BD : 
DC : :  DE :  DF;  and,  by  composition,  BD+DC,  or  BC  :  DE 
4-DF : :  DC  :  DF.  But  BG  being  parallel  to  DF,  DC :  DF ;  : 
BC :  BG,  whence  BC : BG ::  BC: DE+DF.  Therefore  BG 
is  equal  to  DE+DF. 


TEST  EXAMPLES  IN  BOOK  V. 

1.  Through  a  given  point  situated  between  the  sides 
of  an  angle,  to  draw  a  line  terminating  at  the  sides  of 
the  angle,  and  in  such  a  manner  as  to  be  equally  divided 
at  the  point. 

2.  Construct  a  quadrilateral  similar  to  a  given  quadri- 
lateral, the  sides  of  the  latter  having  to  the  sides  of  the 
former  the  ratio  of  2  to  3. 

3.  To  draw  a  line  parallel  to  the  base  of  a  triangle  in  such 
a  manner  as  to  divide  the  triangle  into  two  equal  parts. 

4.  To  construct  a  square  when  the  difference  between 
the  diagonal  and  a  side  is  given. 

5.  Prove  that,  if  a  line  touching  two  circles  cut  another 
line  joining  their  centres,  the  segments  of  the  latter  will 
be  to  each  other  as  the  diameters  of  the  circles. 

6.  Prove  that,  if  from  the  extremities  of  any  chord  in 
a  circle,  perpendiculars  be  drawn  meeting  a  diameter,  the 
points  of  intersection  are  equally  distant  from  the  centre. 

7.  Prove  that,  if  from  the  extremities  of  the  diameter 
of  a  semicircle,  perpendiculars  be  let  fall  on  any  line  cut- 
ting the  semicircle,  the  parts  intercepted  between  those 
perpendiculars  and  the  circumference  are  equal. 

8.  Prove  that  if  two  circles  touch  each  other  externally 
or  internally,  any  straight  line  drawn  through  the  point 
of  contact  will  cut  off  similar  sescments. 


BOOK  V.  117 

9.  To  determine  the  point  in  the  circumference  of  a 
circle  from  which  lines  drawn  to  two  other  given  points 
shall  have  a  given  ratio. 

10.  Prove  that  in  any  right-angled  triangle  the  straight 
line  joining  the  right  angle  and  the  point  of  bisection  of 
the  hypothenuse  is  equal  to  half  the  hypothenuse. 

11.  To  construct  a  polygon  similar  to  a  given  polygon, 
and  bearing  to  it  a  given  ratio. 

12.  Prove  that,  if  from  any  point  wuthin  an  equilateral 
triangle,  perpendiculars  be  drawn  to  the  sides,  they  are 
together  equal  to  a  perpendicular  drawn  from  any  of  the 
angles  to  the  opposite  side. 

13.  Prove  that  if  the  points  of  bisection  of  the  sides  of 
a  given  triangle  be  joined,  the  triangle  so  formed  will  be 
one-fourth  the  given  triangle. 

14.  Prove  that  of  all  triangles  having  the  same  vertic- 
al angle  and  whose  bases  pass  through  a  given  point,  the 
least  is  that  whose  base  is  bisected  in  the  given  point. 

15.  To  bisect  a  given  triangle  by  a  line  drawn  from 
one  of  its  angles. 

16.  To  bisect  a  given  triangle  by  a  line  drawn  from  a 
given  point  in  one  of  its  sides. 

17.  To  determine  a  point  within  a  given  triangle  from 
which  lines  drawn  to  the  several  angles  will  divide  the 
triangle  into  three  equal  parts. 

18.  To  trisect  a  given  triangle  from  a  given  point 
wdthin  it. 

19.  Prove  that  if  the  three  sides  of  a  triangle  be  bi- 
sected, the  perpendiculars  drawn  to  the  sides  at  the  three 
points  of  bisection  will  meet  in  the  same  point. 

20.  Prove  that  if,  from  the  three  angles  of  a  triangle, 
lines  be  drawn  to  the  points  of  bisection  of  the  opposite 
sides,  these  lines  intersect  each  other  in  the  same  point. 

21.  Prove  that  every  trapezium  is  divided  by  its  diag- 
onals into  four  triangles  proportional  to  each  other. 

22.  To  describe  a  triangle  which  shall  be  equal  to  a 
given  equilateral  and  equiangular  pentagon,  and  of  the 
same  altitude. 

23.  Prove  that  if  an  equilateral  triangle  be  inscribed 
in  a  circle,  and  through  the  angular  points  another  be 
circumscribed,  the  inscribed  will  be  one-fourth  the  cir- 
cumscribed. 


118 


ELEMENTS    OF   PXAXE  -GEOMETRY. 


24,  Prove  that  the  ratio  of  the  side  of  a  square  to  the 
diagonal  is  that  of  1  to  the  'v/2. 


A  KEY  TO  THE  TEST  EXAMPLES  IN  BOOK  V. 

1.  A  is  the  given  angle,  and  O  the  given 
point.  Join  AG,  and  produce  it  until  OD 
is  equal  to  AO ;  then  through  D  draw  DF 
parallel  to  AB.  The  line  FE  is  the  line 
required.  The  student  will  have  no  diffi- 
culty in  proving  FO=OE. 

2.  Trisect  the  sides  of  the  former :  two 
of  these  parts  respectively  will  be  the  sides 
of  the  latter :  then  see  Prop.  XIY.,  Bk.  I. 

3.  See  Prop.  XVII.  The  triangle  is  given ;  also  its 
sides.  The  student  will  readily  discover  the  proportion 
whose  fourth  term  will  be  the  portion  of  the  side  through 
the  extremity  of  which  the  line  must  be  drawn  parallel 
to  the  base. 

4.  Let  AB  be  the  given 
difference:  at  the  point  A 
make  DABzz:^  right  angle; 
at  B  make  ABD  1^  right  an- 
gles: produce  these  lines 
until  they  meet :  at  D  make 
BDC=f  right  angle:  pro- 
duce DC  until  it  meets  AB 
prolonged.  Through  C  draw 
EC  parallel  to  AD,  and  AE, 
through  A,  parallel  to  DC. 
The  student  can  now  readily  prove  DE  a  square,  and 
BCrrto  one  side. 

5.  AB    is    tan- 
^^  gent  to  both  cir- 

cles :  CD  joins 
their  centres.  The 
student  will  have 
little  difficulty  in 
proving  the  simi- 
larity of  the  tri- 
andes  ACE  and  EBD. 


BOOK  V. 


119 


6.  At  C  and  D  let  fall  perpendiculars  CE  and  DF, 
meeting  the  diameter  produced.     Required  to  prove  that 


E  and  F  are  equally  distant  from  O.  Draw  OG  perpen- 
dicular to,  and  therefore  bisecting  CD.  By  the  law  of 
similar  triangles  the  student  will  j) 

have  no  difficulty  in  proving  the 
theorem. 

1,  The  demonstration  of  this 
theorem,  by  means  of  the  law  of 
similar  figures,  is  very  simple. 

8.  The  similarity  of  the  triangles  DAC  and  CBE  can 
be  readily  seen.     Hence  the  similarity  of  the  segments. 


9.  A  and  B  are  the  two  given  points.  Join  AB  and 
divide  it  in  D,  so  that  AD :  DB  may 
be  in  the  given  ratio;  bisect  the  arc 
ACB  in  C;  join  CD  and  produce  it 
to  E.  E  is  the  point  required.  Join 
AE  and  EB.  Since  the  arc  AC = the 
arc  CB,  the  angle  AEDrrthe  angle 
BED.     Then  see  Prop.  XXI. 

10.  This   theorem  is  very  simple. 
The  student  has  only  to  inscribe  a  right-angled  triangle 
in  a  circle,  whence  the  proof  becomes  apparent. 

11.  See  Props.  XVII.  and  XXII. 


120 


ELEMENTS   OF   PLANE   GEOMETRY. 


12.  Draw  the  four  perpendiculars 
DE,  DF,  DG,  and  BH. 

Since  triangles  on  the  same  base  or 
equal  bases  are  to  each  other  as  their 
altitudes,  the  triangle  ABC: ADC:: 
BH :  DE ;  also  ABC :  BDC  : :  BH :  DF ; 
also  ABC:ADB::BH:DG;  whence 
ABC  :  ADC  +  BDC+ ADB ::  BH :  DE 
^  +DF+DG.  Since  the  first  term  is 
equal  to  the  second  the  third  is  equal  to  the  fourth. 
Hence  BH=DE+DF+DG. 

13.  If  the  student  bisects  the  sides,  he  will  readily  per- 
ceive the  formation  of  three  parallelograms ;  whence  the 
truth  of  the  theorem  is  easily  established. 

14.  Let  BAC  be  the  vertical  angle  of  any  number  of 
triangles  whose  bases  pass  through  a  given  point,  P ;  and 

BC  be  bisected  in  P ;  ABC 
is  less  than  any  other  tri- 
angle, ADE. 

From  C  draw  CF  paral- 
lel to  AB ;  then  the  angle 
DBP^PCF,  and  the  verti- 
ir-^E  cal  angle  DPB=CPF,  and 
BP=PC.  Therefore  the  triangle  DBP=the  triangle 
PCF;  and  DBP  is  less  than  PCE;  to  each  add  ADPC, 
and  ABC  is  less  than  ADE.  In  the  same  manner  ABC 
may  be  proved  less  than  any  other  triangle  whose  base 
passes  through  P. 

15.  See  Prop.  IX.,  Bk.  II. 
^  16.  Bisect  BC  in  D;  join 

AD  and  PD,  and  from   A 
draw  AE  parallel  to  PD; 
join  PE ;  PE  bisects  ABC. 
The  triangle  ADB = ADC, 
and  proving  the  equality  of 
the    smaller  triangles,  and 
by  subtracting  and  adding 
equals,  the  student  can  easi- 
ly prove  that  APEB^PEC. 
17.  Bisect  AB  and  BC  in  E  and  D ;  join  AD  and  CE ; 
also  B  and  F.     F  is  the  point.     Since  BD=DC,  the  tri- 
anc^les  BAD  and  CAD  are  equal,  and,  for  the  same  reason, 
BDF=DFC.    Therefore  ABF=rAFC.     Ajvain,  since  BE 


BOOK  V. 


121 


=AE,  the  trianorle  BEC  =  the  tri- 
ancrle  AEC;  also  BEFz=AEF. 
Tlierefore  BFC  =  AFC.  Hence 
the  three  parts,  BFC,  BFA,  and 
AFC,  are  equal. 

18.  Trisect  BC  in  D  and  E; 
join  DP  and  PE,  and  from  A  draw 
AF  and  AG  respectively  parallel 
to  them.     Join  PF,  AP,  and  PG. 
These   three    lines    divide   the 
triangle  into  three  equal  parts. 
Join  AD  and  AE.     Since  AF 
is  parallel  to  PD,  the  triangle 
APF  =  ADF;  to  each  of  them 
addABF;  and  ABFP=:ADB. 
In  the  same  manner  ACGP= 
AEC.      Hence     FPG  =  ADE. 
But  the  triangles  ABD,  AEC, 
and  ADE  are  equal,  each  hav- 
ing an  equal  base  and  altitude. 
Therefore  ABFP,  ACGP,  and  FPG  are  also  equal. 

19.  Let  the  sides  be  bisected  in  D,  E,  and  F.     Draw  the 
perpendiculars  EG  and  GF,  -^ 
meeting  in  G.     The  perpen- 
dicular   at    D    also    passes 
through  G.     Join  GD,  GA, 

GB,  and  GC.     Since   AF=                  j^y^     /  \e 
FC,  and  FG  is  common,  and 
the  angles  at  F  right  angles, 
AG=:GC.    In  the  same  man- 
ner it  may  be  shown  that  GC    — —  ^ 

=rGB.     Therefore  AG  =.GB.  A  1'  C 

But  AD =DB,  and  DG  is  common  to  the  triangles  ADG  and 
BDG.  Hence  the  angles  at  D  are  equal,  and  each  must  be 
a  right  angle ;  or  the  perpendicular  at  D  passes  through  G. 

20.  Let  the  sides  be  bisected  in  D,  E,  and  F.  Join  AE, 
CD,  meeting  each  other  in  G.  Join  BG  and  GF ;  BGF 
is  a  straight  line.  Join  EF,  meeting  CD  in  H.  Then  FE 
is  parallel  to  AB,  and  therefore  the  triangles  DAG  and 
GEH  are  equiangular.     Hence 

^DA:DG::HE:HG; 
or  DB:DG::HF:HG; 

F 


122 


ELEMENTS  OF  PLAXE  GEOMETRY. 


that  is,  the  sides  about  the  equal 
angles  are  proportional.  Hence 
the  triangles  BDG  and  GHF  are 
similar;  and  the  angle  DGB=: 
HGF.  Therefore  BG  and  GF 
are  in  the  same  straight  line. 

21.  See  Prop.  XIII. 

22.  ABODE  is  the  given  penta- 
gon. Produce  CD  both  ways; 
join  AC  and  AD.  Through  B  and 
E  draw  BG  and  EF  parallel  to 

AC  and  AD,  each  to 
each.     Join  AG  and 
AF.     AGF  is  the  re- 
quired triangle.    The 
student   can   readily 
perceive  that,  by  add- 
ing certain  equal  tri- 
angles to  a  common 
triangle,  the  truth  of 
the  problem  will  be 
established. 
23.  ABC  is  the  equilateral 
triangle  inscribed ;  and  DICFis 
the  equilateral  circumscribed. 
Prove  that  ABC  is  one-fourth 
ofDEF. 

The  angle  DAB  =  ACB 
(Prop.  XII.,  Bk.  III.).  But 
ACBzizABC.  Therefore  DAB 
rrABC;  these  are  alternate 
angles.  Hence  DE  is  parallel 
to  BC.  In  the  same  manner  it 
may  be  shown  that  DF  is  par- 
allel to  AC.  Therefore  ACBD  is  a  parallelogram ;  and 
the  triangle  ABC— ABD;  and  in  like  manner  it  can  be 
proved  tliat  ABC^rAEC  or  BCF.  Hence  ABC  =  one- 
fourth  of  DEF. 

24.  ABCD  is  the  square,  and  AC  its  diagonal.  From 
the  diagonal  cut  off  AF=AB;  the  remainder,  CF, 
must  be  compared  with  CB.  Join  FB,  and  draw  FE 
perpendicular    to   AC.      The    angle    ABE  =  AFE;    and 


IU)()K    V. 


123 


ABFz=AFB;  and  from  tlie  former  equals  subtract  the 

O C      latter,  and  EFJ>  remains  =  to  EBF. 

Hence  tlie  triangle  13EF  is  isosceles, 
and  BErrEF.  the  angle  FCE  is  half 
a  right  angle.  But  CFE  is  a  right  an- 
gle.' Therefore  FEC  is  half  a  right  an- 
gle. Hence  FC=:FE;  and  CE  is  the 
diagonal  of  a  square  whose  side  is  FC. 
Hence,  after  CF  has  been  taken  from 
CB,  it  remains  to  take  CF  from  CE ;  ^ 
that  is,  to  compare  the  side  of  a  square  with  its  diagonal : 
and  we  shall  find  precisely  the  same  difficulty  in  the 
next  step  of  the  process ;  so  that,  continue  as  far  as  we 
please,  we  shall  never  arrive  at  a  term  in  which  there 
will  be  no  remainder.  Therefore  there  is  no  common 
measure  for  the  diagonal  and  a  side  of  a  square.  If  the 
side  of  the  square  be  represented  by  1,  the  diagonal  will 
be  the  square  root  of  2.  But  this  is  only  an  approxi- 
mate value. 

[The  teacher  should  be  very  careful  not  to  show  the  pupils  too  much 
in  the  study  of  the  Test  Examples ;  neither  should  they  be  permitted 
to  consume  too  much  time  in  vain  efforts  to  demonstrate  or  solve  ques- 
tions too  difficult  for  their  comprehension.  The  teacher  should  exercise 
a  nice  discretion.] 


APPENDIX. 


IVIENSURATION  OF  SURFACES. 

1.  The  area  of  a  figure  is  the  measure  of  its  surface. 

2.  A  square  whose  side  is  one  inch,  one  foot,  or  one  yard,  is  called 
the  unit  of  measure^  and  the  area  of  any  figure  is  computed  by  the  num- 
ber of  those  squares  contained  in  that  figure. 

Problem  I. 

To  find  the  area  of  a  parallel  ogr  am,  whether  it  he  a  square,  a  rectan- 
gle, a  rhombus^  or  a  rhomboid. 

Rule. — Multiply  the  length  by  the  perpendicular  height,  and  the  prod- 
uct will  be  the  area.* 

Examples. 

1.  To  find  the  area  of  a  parallelogram  whose  length  is  12  feet  3  inches, 
and  height  8  feet  6  inches : 

3  inches  =  .25  foot  and  6  inches. 5  foot. 
Then    •  12.25  feetX8. 5  feet =104. 125  feet  area. 

2.  Find  the  area  of  a  square  whose  side  is  35.25  chains,  t 

Ans.  124  acres  1  rood  1  pole. 

3.  Find  the  area  of  a  rectangular  board  whose  length  is  12 J  feet,  and 
breadth  9  inches.  Ans.  9f  feet. 

4.  To  find  the  content  of  a  piece  of  land  in  the  fonii  of  a  rhombus,  its 
length  being  6.2  chains,  and  height  5.45  chains. 

Ans.  3  acres  1  rood  20  poles. 

5.  Required  the  area  of  a  rhomboid  whose  length  is  10.51  chains,  and 
breadth  4.28  chains.  Ans.  4  acres  1  rood  39.7  poles. 

Problem  II. 

To  find  the  area  of  a  triangle  when  the  base  and  perpendicular  height 
are  given. 

Rule. — Multiply  the  base  by  the  perpendicular  height,  and  take  half 
the  product ;  or,  multiply  the  base  by  half  the  perpendicular  height,  or  the 
perpendicular  height  by  half  the  base.X 

*  See  Proposition  II.,  Bk.  V.,  Scholium. 

t  The  student  must  know  the  Table  of  Measures. 

X  See  Proposition  II.,  Bk.  V.,  Cor. 


120 


APPENDIX. 


Examples. 
1.  Find  the  area  of  a  triangle  whose  base  is  49  feet,  and  height  25| 

feet. 


49X25.20 


=G18.625  square  feet. 


2.  How  many  square  yards  in  the  triangle  whose  base  is  40  feet,  and 
perpendicular  height  30  ?  Ans.  66|  square  yards. 

3.  To  find  the  area  of  a  triangle  whose  base  is  18  feet  4  inches,  and 
height  11  feet  10  inches.  Ans.  108  feet  5|  inches. 

4.  Required  the  area  of  a  triangle  whose  base  is  12.25  chains,  and 
height  8.5  chains.  Ans.  5  acres  0  rood  36  poles. 

Problem  III. 
To  find  the  area  of  a  triangle  whose  three  sides  only  are  given. 
Rule. — From  half  the  sum  of  the  three  sides  subtract  each  side  sepa- 
rately :  multiply  the  half  sum  and  the  three  remainders  continually  to- 
get  her,  and  the  square  root  of  the  product  will  be  the  area  required.* 


Demonstration.— Let 


AC=c,  CB=(f,  andABrr&;  and 
c2=:AD2-f  DC2 ; 
d2-DB2+DC2 ; 
c2— d2-AD2— DB2 ; 
(c+cO  (c-d)=(AD+DB)  (AD— DB) ; 
(c+d)  {e—d)=b{AB—DB)  from  Fig.  1 ; 
(c+d)  (c— d)=:&(AD+DB)  from  Fig.  2  ; 

c  c 


AB-DbJ^±^^^,  and  AD+DbJ^^^J^-^; 

AD-DB=:^— -1-,  aud  AD+DB=:— j— ; 
b  0 

C2 ^2      5 

adding  half  to  half  diff.,  —^ — 1--=AD,  the  greater  segment 


2& 
CD 


=,-,^.=J.-C2=^)\ 


APPENDIX.  127 

Examples. 

1.  Required  the  area  of  a  triangle  whose  three  sides  are  24,  36,  and 

48  chains  respectively : 

24+36+48     108     „.      i  •  .   .    ,.     i 
^T =-—==54,  which  IS  the  ^  sum. 

Then  54-24=30,  first  diff.  ;  54-36  =  18,  second  diff.  ;  and  54-48  = 
6,  third  diff. 

(Multiplying),  54 X 30X18 X 6=174960  ; 

(Extracting  square  root),  \/ 174960=418.282,  area  required. 

2.  Required  the  area  of  a  triangle  whose  three  sides  are  13, 14,  and  15 
feet.  J-Tw.  84  sq.  feet. 

3.  How  many  acres  are  there  in  a  triangle  wliose  three  sides  are  49, 
50.25,  and  25.69  chains?  Ans.  61.498  acres. 

4.  Required  the  area  of  a  right-angled  triangle  whose  hypothenuse  is 
SO,  and  the  other  two  sides  30  and  40.  Ans.  600. 

5.  Required  the  area  of  an  equilateral  triangle  whose  side  is  25. 

Ans.  270.6328. 

6.  Required  the  area  of  an  isosceles  triangle  whose  base  is  20,  and 
each  of  its  sides  15.  Ans.  111.803. 

7.  Required  the  area  of  a  triangle  whose  three  sides  are  20,  30,  and 
40  chains.  Ans.  2^  acres  7  poles. 

Problem  IV. 

Any  two  sides  of  a  right-angled  triangle  being  given  to  find  the  third 
side. 

Rule. — When  two  sides  are  given  to  find  the  hypothenuse.*  Add  the 
squares  of  the  two  sides  together  and  extract  the  square  root  of  the  sum. 

When  the  hypothenuse  and  one  side  are  given,  to  find  the  other.  From 
the  square  of  the  hypothenuse  subtract  the  square  of  the  given  side,  and  the 
square  root  of  the  remainder  will  be  the  required  side. 

56'*+33''=3136+1089=4225;  and  -/ 4225 =65,  ^ns. 


^p^VI(H-c)2-d2]  [da_(6_c)2]^ 


26 


area  _V('>+c+'i)  jb+c-d)  jb-c+d)  {-b+e+d)_ 
Area  =  {5±|Mx?±f=^x?=|t^x:±t£±£}* 

'  [In  this  demonstration  the  student  must  understand  very  clearly  the  principles  of  factoring  in  Alge- 
bra.] 

♦  See  Prop.  XIII.,  Bk.  II. 


128'  APPENDIX.      ' 

1.  Given  the  base  of  a  riglit-angled  triangle  5G,  and  the  perpendicular 
33,  what  is  the  hypothenuse  ? 

2.  If  the  hypothenuse  be  53,  and  the  base  45,  what  is  the  perpendic- 
ular? Ans.  28. 

3.  The  base  of  a  right-angled  triangle  is  77,  and  the  peipendicular  36, 
what  is  the  hypothenuse  ?  Ans.  85. 

4.  The  hypothenuse  of  a  right-angled  triangle  is  109,  and  the  perpen- 
dicular 60,  what  is  the  base  ?  Ans.  i)l. 

5.  The  height  of  a  precipice  standing  close  by  the  side, of  a  river  is 
103  feet,  and  a  line  of  320  feet  will  reach  from  the  top  of  it  to  the  oppo- 
site bank:  required  the  width  of  the  river.  u4.ns.  302.97  feet. 

6.  A  ladder  50  feet  long  being  placed  in  a  street  reached  a  window 
28  feet  from  the  ground  on  one  side ;  and  by  turning  it  over,  without  re- 
moving the  foot,  it  reached  another  window  36  feet  high  on  the  other 
side :  required  the  width  of  the  street.  Ans.  23.3238  feet. 

Problem  V. 
To  find  the  area  of  a  trapezium. 
Rule. — Divide  the  trapezium  {an  irregular  quadrilateral)  into  two  tri- 
angles hy  a  diagonal;  then  to  this  diagonal  let  perpendiculars  fall  from 
the  opposite  angles.     Find  the  area  of  each  triangle  hy  Problem  II. 

1.  Find  the  area  of  a  trapezium  whose  diagonal  is  84,  and  the  two 
pei*pendiculars  21  and  28  respectively. 

(21  +  28)  X  84=49X84=4116;  and  ^^=2058,  u4«s. 

2.  Required  the  area  of  a  trapezium  whose  diagonal  is  80. 5,  and  the 
two  perpendiculars  24.5  and  30.1.  ^ws.  2197.65. 

3.  What  is  the  area  of  a  trapezium  whose  diagonal  is  108  feet  6 
inches,  and  the  perpendiculars  56  feet  3  inches  and  60  feet  9  inches. 

Ans.  6347  feet  3  inches. 

4.  How  many  square  yards  of  paving  are  there  in  the  trapezium  whose 
diagonal  is  65  feet,  and  the  two  perpendiculars  let  fall  on  it  28  and  33j 
feet.  Ans.  2223^5  yards. 

Problem  VI. 
To  find  the  area  of  a  trapezoid. 
'Rult:.— Multiply  the  sum  of  the  parallel  sides  by  the  perpendicular  dis- 
tance between  them,  and  half  the  product  will  be  the  area.* 

1.  The  parallel  sides  of  a  trapezoid  are  750  and  1225,  and  the  perpen- 
dicular distance  between  them  1540  links :  required  the  area. 

1  ^^40 
(1225  +  750)  X— — =1975  X  770  =  152075  links =15  acres  33  poles,  Ans. 

2.  How  many  square  feet  are  contained  in  the  plank  whose  length  is 

*  This  is  simply  finding  the  areas  of  the  two  triangles. 


APPENDIX. 


129 


1*2  feet  6  inches,  the  breadth  at  the  greater  end  being  15  inches,  and  at 
the  less  end  11  inches.  Ans.  13  J  J  feet. 

3.  The  parallel  sides  of  a  trapezoid  are  12.41  and  8.22  chains,  and 
the  perpendicular  distance  5.15  chains :  required  the  area. 

Ans.  5  acres  120.995  poles. 

Problem  VII. 

To  find  the  area  of  an  irregular  pohjgon. 

Rule. — Divide  the  polygon  into  triangles  and  trapeziums;  then  find 
the  area  of  all  these  {by  preceding  problems^  and  their  sum  will  be  the 
area  required. 

1.  Find  the  area  of  the  irregular  figure  ABCDEFG,  in  which  are 
given  the  following  diagonals  and  perpendiculars:  AC =55,  rD=52, 
GC=54,  Gm=13,  B7i=18,  Go=12,  Ep=S,  and  D7=23. 

Ans.  1878^. 


Problem  VIII. 

To  find  the  area  of  a  regular  polygon. 

Rule. — Multiply  the  perimeter  of  the  polygon^  or  sum  of  its  sides,  by 
the  perpendicular  drawn  from  its  centre  on  one  of  its  sides,  and  take  half 
*he  product  for  the  area.* 

1.  Find  the  area  of  a  regular  pentagon,  each  of  whose  sides  is  25  feet, 
and  the  perpendicular  from  the  centre  on  each  side  17.2047737. 

25X5=125,  the  perimeter, 
and  125  x  17.2047737=2150.5967125. 

The  half  of  this  is  1075.298356,  the  area  required. 

2.  Required  the  area  of  a  hexagon  whose  side  is  14.6  feet,  and  perpen- 
dicular 12.64.  Ans.  553.632  feet. 

*  This  is  simply  resolving  the  polygon  into  as  many  equal  triangles  as  tht 
fio'ure  has  sides,  finding  the  area  of  each,  and  taking  the  sum. 

F2 


1  :I0 


APPEXDIX. 


o.  Find  the  area  of  a  heptagon  whose  side  is  19.38,  and  perpendicular 
20.  Ans.  1356.6. 

Problem  IX. 
To  find  the  area  of  a  regular  polygon  when  a  side  only  is  given. 
Rule. — Multiply  the  square  of  the  side  by  the  number  standing  oppo- 
site to  its  name  in  the  following  table,  and  the  product  will  be  the  area."^ 


No.  of 
sides. 


Names. 


Multipliers. 


3. 

4. 
5. 
6. 

7. 

8. 

9. 
10. 
11. 
12. 


Trigon,  or  equilateral  A. 

Square 

Pentagon , 

Hexagon  

Heptagon 

Octagon 

Nonagon 

Decagon 

Undecagon 

Duodecagon 


0.433013- 
1.000000 
1.720477  + 
2.598076  + 
3.633912  + 
4.828427  + 
6.181824  + 
7.694209- 
9.365640- 
11.196152- 


1.  Find  the  area  of  a  regular  pentagon  whose  side  is  25  feet. 

25^^=625. 
The  tabular  area  is  1.720477. 

625X1.720477  =  1075.298375,  Ans. 

2.  Find  the  area  of  an  equilateral  triangle  whose  side  is  20. 

Ans.  173.20508. 

3.  Find  the  area  of  a  hexagon  whose  side  is  20. 

Ans.  1039.23048. 

4.  Find  the  area  of  an  octagon  whose  side  is  16. 

Ans.  1236.0773. 

5.  Find  the  area  of  a  duodecagon  whose  side  is  125. 

Ans.  174939.875. 
Problem  X. 
The  diameter  of  a  circle  being  given  to  find  the  circumference,  or  the 
circumference  being  given  to  find  the  diameter. 

Rule. — Multiply  the  diameter  by  3.1416,  and  the  product  will  be  the 
circumference;  or,  divide  the^  circumference  by  3.1416,  and  the  quotient 
will  be  the  diameter.^ 

*  The  multipliers  in  the  table  are  the  areas  of  the  polygons  to  which  they  be- 
long when  the  side  is  unity,  or  1.  All  similar  figures  are  to  each  other  as  the 
squares  of  their  like  sides.  See  Prop.  XVII,,  Bk.  V.  Hence  1^ :  given  side 
squared  : :  given  area  :  area  required  ;  or  given  side  squared  X  given  area  = 
area  required. 

t  The  proportion  of  the  diameter  to  the  circumference  has  never  yet  been  ex- 
actly obtained:  nor  can  a  square  or  other  straight-lined  figure  be  found  that 
shall  be  equal  to  a  given  circle.  This  is  the  celebrated  problem  called  squaring 
the  circle,  which  has  never  been  solved. 


Arr::M)ix.  131 

Also,  as       7  :    22  : :  the  diameter  :  the  circumference  ; 
or,  as  22  :      7  : :  the  circumference  :  the  diameter. 

As  113  :  355  : :  the  diameter  :  the  circumference ; 

or,  as  355  :  113  : :  the  circumference  :  the  diameter. 

1.  Find  the  circumference  of  a  circle  whose  diameter  is  20. 

As  7  :  22  : :  20  :  the  circumference, 

2*>X20 
tZ±£lL=G2^-,  Ans. 

7  ' 

2.  If  the  circumference  of  a  circle  is  354,  what  is  the  diameter? 

A71S.  112.G81. 

3.  If  the  circumference  of  the  earth  be  25,000  miles,  what  is  the  diam- 
eter ?  Ans,  7958,  nearly. 

4.  What  is  the  circumference  of  a  circle  whose  diameter  is  40  feet  ? 

Atis.  125.6010. 

PllOBLEM   XI. 

To  find  the  area  of  a  circle. 

Rule. — 1.  Multiply  half  the  circumference  hij  half  the  diameter* 

2.  Square  the  diameter^  and  multiply  that  square  by  the  decimal .  7854.  t 

3.  Square  the  circumference,  and  multiply  that  square  by  the  decimal 
.07958. 

1.  Find  tlie  area  of  a  circle  whose  diameter  is  10,  and  its  circumfer- 
ence 31.41G. 

r>y  Rule  1.  By  Rule  2. 

•"^•^^^^xl?  =  78.54.  10^X.7854  =  78.5t. 


2 

By  Rule  3. 
31. 416' X. 07958  =  78.54. 

2.  Find  the  area  of  a  circle  whose  diameter  is  7,  and  circumference  22. 

Ans.  38 J. 

3.  How  many  square  yards  in  a  circle  whose  diameter  is  3J^  feet  ? 

Ans.  1.069. 

4.  Find  the  area  of  a  circle  whose  circumference  is  12  feet. 

Ans.  11.4595. 

5.  How  many  square  feet  are  there  in  a  circle  whose  circumference  is 
10.9956?  Ans.  86.5933. 

Problem  XII. 

To  find  the  area  of  a  circular  ring,  or  of  the  space  included  between 

*  Tlnle  L— In  this  rule  the  circumference  is  made  np  of  an  infinite  number  of 
straight  lines,  each  of  which  becomes  the  base  of  a  trianfjlc  whose  perpendicular 
is  the  radius,  or  half  the  diameter.  Heuce  \  the  circumference  X  k  the  diameter 
would  be  the  area  of  an  infinite  number  of  triangles,  or  the  area  of  the  circle. 

t  Rule  Il.-^Iu  this  rule  the  area  of  a  circle  whose  diameter  is  1=.7S54 ;  and, 
by  the  law  of  similar  figures,  1^ :  diam.'  : :  .TS54 :  area  required. 


132  Xppexdix. 

the  circumferences  of  two  circles,  the  one   being  contained  within   the 
other. 

Rule. — Find  the  areas  of  the  two  circles  by  the  last  problem^  and  sub- 
tract the  less  from  th»  greater. 

1,  The  diameters  of  two  conceRtric  circles  being  10  and  6 :  required 
the  area  of  the  ring  contained  between  their  circumferences. 

Ans.  50.2G56. 

2.  What  is  the  ai*ea  of  the  ring  the  diameters  of  whose  bounding  cir- 
cles ai*e  10  and  20  ?  Ans.  235.62. 


PRACTICAL  EXAMPLES. 

1.  What  will  the  glazing  of  a  triangular  sky-light  come  to  at  20  cents 
per  foot,  the  base  being  12  feet  6  inches,  and  the  perpendicular  height 
6  feet  9  inches  ?  Ans.  ^16.875. 

2.  From  a  mahogany  plank  26  inches  broad,  a  yard  and  a  half  is  to 
be  sawed  off;  what  distance  from  the  end  must  the  line  be  struck  ? 

Ans.  6.23  feet. 

3.  How  many  square  yards  in  a  ceiling  which  is  43  feet  3  inches  long 
and  25  feet  6  inches  broad  ?  An^.  122J. 

4.  What  is  a  marble  slab  worth  whose  length  is  5  feet  7  inches,  and 
breadth  1  foot  10  inches,  at  80  cents  per  foot  ?  Ans.  $8,188. 

5.  What  is  the  difference  between  a  floor  48  feet  long  and  30  feet 
broad,  and  two  others  each  of  half  the  dimensions  ?         Ans.  720  feet. 

6.  The  two  sides  of  an  obtuse-angled  triangle  are  20  and  40  poles ; 
what  must  be  the  length  of  the  third  side,  so  that  the  triangle  may  con- 
tain just  an  acre?  Ans.  58.876  poles. 

7.  A  circular  fish-pond  is  to  be  dug  in  a  garden  that  shall  take  up 
just  half  an  acre ;  what  must  the  length  of  the  cord  be  that  describes 
the  circle?  ^ as.  27.75  yards. 

8.  Suppose  a  ladder  100  feet  long  placed  against  a  pei-pendicular  wall 
100  feet  high  ;  how  far  would  the  top  of  the  ladder  move  down  by  pull- 
ing out  the  bottom  thereof  10  feet  ?  Ans.  .5012563  foot. 

9.  A  telegraph  pole  was  so  nearly  broken  through  by  a  blast  of  wind 
that  the  top  fell  to  the  ground  15  feet  from  the  base  of  the  pole :  what 
was  the  height  of  the  whole  telegraph  pole,  supposing  the  length  of  the 
piece  that  fell  to  be  39  feet  ?  Ans.  75  feet. 

10.  Three  persons,  whose  residences  are  at  the  vertices  of  a  triangular 
area,  the  sides  of  which  are  severally  10,  11,  and  12  chains,  wish  to  dig 
a  well  which  shall  be  at  the  same  distance  from  the  residence  of  each. 
Find  the  point  for  the  well,  and  its  distance  from  their  residences. 

Ans.  6.405  chains. 

THE  END. 


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